2.1 Einführung zur Integralrechnung

Aus Online Mathematik Brückenkurs 2

(Unterschied zwischen Versionen)
Wechseln zu: Navigation, Suche
K (Regenerate images and tabs)
Zeile 2: Zeile 2:
{| border="0" cellspacing="0" cellpadding="0" height="30" width="100%"
{| border="0" cellspacing="0" cellpadding="0" height="30" width="100%"
| style="border-bottom:1px solid #797979" width="5px" |  
| style="border-bottom:1px solid #797979" width="5px" |  
-
{{Vald flik|[[2.1 Inledning till integraler|Teori]]}}
+
{{Vald flik|[[2.1 Inledning till integraler|Theory]]}}
-
{{Ej vald flik|[[2.1 Övningar|Övningar]]}}
+
{{Ej vald flik|[[2.1 Övningar|Exercises]]}}
| style="border-bottom:1px solid #797979" width="100%"|  
| style="border-bottom:1px solid #797979" width="100%"|  
|}
|}
{{Info|
{{Info|
-
'''Innehåll:'''
+
'''Content:'''
-
* Integralens definition (översiktligt).
+
*Definition of an integral (overview).
-
* Integralkalkylens huvudsats.
+
* Fundamental theorm of integral calculus.
-
* Primitiv funktion till <math>x^\alpha</math>, <math>1/x</math>, <math>e^x</math>, <math>\cos x</math> och <math>\sin x</math>.
+
* Primitive function for <math>x^\alpha</math>, <math>1/x</math>, <math>e^x</math>, <math>\cos x</math> and <math>\sin x</math>.
-
* Primitiv funktion till summa och differens.
+
* Primitive function for sum and difference.
}}
}}
{{Info|
{{Info|
-
'''Lärandemål:'''
+
'''Learning outcomes:'''
-
Efter detta avsnitt ska du ha lärt dig att:
+
After this section, you will have learned to :
-
* Tolka integraler som areor, dvs. "area ovanför <math>x</math>-axeln" minus "area under <math>x</math>-axeln".
+
* Interpret integrals as signed areas, that is, " the area above the <math>x</math>-axis" minus " area below the <math>x</math>-axis".
-
* Förstå andra tolkningar av integralen, t.&nbsp;ex. densitet/massa, fart/sträcka, ström/laddning, etc.
+
* Understand other interpretations of the integral, for example. density / mass, speed / displacement, power / charge , etc.
-
* Kunna bestämma primitiv funktion till <math>x^\alpha</math>, <math>1/x</math>, <math>e^{kx}</math>, <math>\cos kx</math>, <math>\sin kx</math> och summa/differens av sådana termer.
+
* Determine primitive function <math>x^\alpha</math>, <math>1/x</math>, <math>e^{kx}</math>, <math>\cos kx</math>, <math>\sin kx</math> and the sum / difference of such terms.
-
* Kunna räkna ut area under en funktionskurva.
+
*Calculate the area below the curve of a function.
-
* Kunna räkna ut area mellan två funktionskurvor.
+
* Calculate the area between two curves of two functions.
-
* Veta att alla funktioner inte har primitiv funktion som kan skrivas som ett analytiskt slutet uttryck, t.ex. <math>e^{x^2} </math>, <math>(\sin x)/x</math>, <math>\sin \sin x</math>, etc.
+
* Recognise that all functions do not have primitive functions that can be written as a closed analytical expression, such as <math>e^{x^2} </math>, <math>(\sin x)/x</math>, <math>\sin \sin x</math>, etc.
}}
}}
-
== Area under en funktionskurva ==
+
== Area below the curve of a function ==
-
Vi har tidigare sett att lutningen på en funktionskurva är intressant. Den ger oss information om hur funktionen ändras och har stor betydelse i många tillämpningar. På ett liknande sätt är den area som bildas mellan en funktionskurva och ''x''-axeln betydelsefull. Den är naturligtvis beroende av funktionskurvans utseende och därmed intimt besläktad med funktionen i fråga. Det är lätt att inse att denna area har en praktisk betydelse i många olika sammanhang.
+
We previously have found that the slope of a curve of a function is interesting. It gives us information about how the function changes and has great significance in many applications. In a similar way the area between the curve of a function and the ''x''-axis is of importance. It of course is dependent on the curves appearance and thus closely related to the function in question. It is easy to see that this area has practical significance in many different contexts.
-
Om ett föremål rör sig så kan vi beskriva dess hastighet ''v'' efter tiden ''t'' i ett ''v-t''-diagram. Vi ser här tre olika fiktiva exempel:
+
If an object is moving, we can illustrate its speed ''v'' plotted against time ''t'' in a ''v,t''-diagram. We can see in the figure below three different hypothetical examples:
Zeile 45: Zeile 45:
|-
|-
||
||
-
| valign="top" |<small>Föremålet rör sig med den konstanta farten 5.</small>
+
| valign="top" |<small> The object moves at a constant speed of 5.</small>
||
||
-
| valign="top" |<small>Föremålet rör sig med den konstanta farten 4 för att vid en stöt när ''t''&nbsp;=&nbsp;3 plötsligt öka farten till 6.</small>
+
| valign="top" |<small> The object moves at a steady speed of 4 when an impact at ''t''&nbsp;=&nbsp;3 suddenly increases the speed to 6.</small>
||
||
-
| valign="top" |<small>Föremålet glider ner för ett sluttande plan och har en linjärt ökande fart.</small>
+
| valign="top" |<small>The object is sliding down a sloping plane and has a linearly increasing speed. </small>
||
||
|}
|}
-
Den tillryggalagda sträckan är i respektive fall
+
The distance travelled is in each case is
{{Fristående formel||<math>s(6) = 5\cdot 6 = 30\,\mbox{m},\quad
{{Fristående formel||<math>s(6) = 5\cdot 6 = 30\,\mbox{m},\quad
Zeile 60: Zeile 60:
s(6) = \frac{6\cdot 6}{2} = 18\,\mbox{m}\,\mbox{.}</math>}}
s(6) = \frac{6\cdot 6}{2} = 18\,\mbox{m}\,\mbox{.}</math>}}
-
I samtliga fall ser man att föremålets tillryggalagda sträcka motsvaras av arean under funktionskurvan.
+
In each cases, you see that the distance travelled by the object is matched by the area below the curve.
-
Fler exempel på vad arean under en funktionskurva kan symbolisera följer nedan.
+
More examples of what the area below a curve can symbolise are shown below.
<div class="exempel">
<div class="exempel">
-
'''Exempel 1'''
+
''' Example 1'''
Zeile 78: Zeile 78:
|-
|-
||
||
-
| valign="top" |<small>En solcell som bestrålas av ljus med en viss effekt ''p'' kommer ha mottagit en energi som är proportionell mot arean under grafen ovan.</small>
+
| valign="top" |<small> A solar cell which has been exposed to light of power p will have received energy that is proportional to the area under the above graph. </small>
||
||
-
| valign="top" |<small>Kraften ''F'' som verkar i ett föremåls rörelseriktning utför ett arbete som är proportionell mot arean under grafen ovan.</small>
+
| valign="top" |<small>The force ''F'' applied to an object along the direction of its motion does work that is proportional to the area under the above graph. </small>
||
||
-
| valign="top" |<small>En kondensator som laddas upp med en ström ''i'' kommer ha en laddning som är proportionell mot arean under grafen ovan.</small>
+
| valign="top" |<small> A capacitor that is charged by a current ''i'' will receive a charge which is proportional to the area under the above graph. </small>
||
||
|}
|}
Zeile 89: Zeile 89:
-
== Integralbeteckningen ==
+
== The notation for an integral. ==
-
För att beskriva arean under en funktionskurva i symbolform inför man ''integraltecknet'' <math>\,\smallint\,</math> och gör följande definition:
+
In order to describe the area below the curve of a function in symbolic form one introduces the ''integral sign'' <math>\,\smallint\,</math> :
<div class="tips">
<div class="tips">
-
Med integralen av den positiva funktionen <math>f(x)</math> från <math>a</math> till <math>b</math> menas arean mellan kurvan <math>y=f(x)</math> och ''x''-axeln från <math>x=a</math> till <math>x=b</math> , vilket med symboler skrivs
+
The integral of a positive function <math>f(x)</math> from <math>a</math> to <math>b</math> is understood to mean the area between the curve <math>y=f(x)</math> and the interval of the ''x''-axis between <math>x=a</math> and <math>x=b</math> , and is written with the notation
{{Fristående formel||<math>\int_{a}^{\,b} f(x)\, dx\,\mbox{.}</math>}}
{{Fristående formel||<math>\int_{a}^{\,b} f(x)\, dx\,\mbox{.}</math>}}
-
Talen <math>a</math> och <math>b</math> kallas undre respektive övre integrationsgräns, <math>f(x)</math> kallas integrand och <math>x</math> integrationsvariabel.
+
The numbers <math>a</math> and <math>b</math> are called the lower and upper limits of integration respectively, <math>f(x)</math> is called the integrand and <math>x</math> the variable of integration.
</div>
</div>
<div class="exempel">
<div class="exempel">
-
'''Exempel 2'''
+
''' Example 2'''
{| width="100%"
{| width="100%"
-
| width="95%" |Arean under kurvan <math>y=f(x)</math> från <math>x=a</math> till <math>x=c</math> är lika med arean från <math>x=a</math> till <math>x=b</math> plus arean från <math>x=b</math> till <math>x=c</math>. Detta betyder att
+
| width="95%" | The area below the curve <math>y=f(x)</math> from <math>x=a</math> to <math>x=c</math> is equal to the area from <math>x=a</math> to <math>x=b</math> plus the area from <math>x=b</math> to <math>x=c</math>. This means that
{{Fristående formel||<math>\int_{a}^{\,b} f(x)\, dx + \int_{b}^{\,c} f(x)\, dx
{{Fristående formel||<math>\int_{a}^{\,b} f(x)\, dx + \int_{b}^{\,c} f(x)\, dx
Zeile 118: Zeile 118:
<div class="exempel">
<div class="exempel">
-
'''Exempel 3'''
+
''' Example 3'''
{| width="100%"
{| width="100%"
-
| width="95%" | För ett föremål, vars hastighet förändras enligt funktionen <math>v(t)</math> kan den tillryggalagda sträckan efter 10&nbsp;s beskrivas med integralen
+
| width="95%" | For an object, whose speed is changing according to the function <math>v(t)</math> the distance travelled after 10&nbsp;s is characterised by the integral
{{Fristående formel||<math>s(10) = \int_{0}^{10} v(t)\, dt\,\mbox{.}</math>}}
{{Fristående formel||<math>s(10) = \int_{0}^{10} v(t)\, dt\,\mbox{.}</math>}}
-
''Anm.'' Vi antar att hastigheten och sträckan mäts med samma längdenhet.
+
''Note .'' We assume that speed and distance are measured using the same units of length.
 +
 
| width="5%" |
| width="5%" |
||{{:2.1 - Figur - Area s(10) i ett v-t-diagram}}
||{{:2.1 - Figur - Area s(10) i ett v-t-diagram}}
Zeile 131: Zeile 132:
<div class="exempel">
<div class="exempel">
-
'''Exempel 4'''
+
''' Example 4'''
-
Vatten rinner in i en tank med en hastighet som är <math>f(t)</math>&nbsp;liter/s efter <math>t</math> sekunder. Integralen
+
Water is flowing into a tank at a rate of <math>f(t)</math>&nbsp;liter/s at the time <math>t</math>. The integral
{{Fristående formel||<math>\int_{9}^{10} f(t)\, dt</math>}}
{{Fristående formel||<math>\int_{9}^{10} f(t)\, dt</math>}}
-
anger då hur många liter som rinner in i tanken under den tionde sekunden.
+
specifies the amount in litres which flows into the tank during the tenth second.
</div>
</div>
<div class="exempel">
<div class="exempel">
-
'''Exempel 5'''
+
''' Example 5'''
-
Beräkna integralerna
+
Calculate the integrals
{| width="100%"
{| width="100%"
Zeile 150: Zeile 151:
<ol type="a">
<ol type="a">
<li> <math>\int_{0}^{4} 3 \, dx</math><br> <br>
<li> <math>\int_{0}^{4} 3 \, dx</math><br> <br>
-
Integralen kan tolkas som arean under kurvan (linjen) <math>y=3</math>
+
The integral can be interpreted as area below the curve (the line) <math>y=3</math>
-
från <math>x = 0</math> till <math>x = 4</math>,
+
going from <math>x = 0</math> to <math>x = 4</math>,
-
dvs. en rektangel med basen 4 och höjden 3, <br>
+
i.e. a rectangle with the base 4 and height 3, <br>
<center><math>\int_{0}^{4} 3 \, dx = 4 \cdot 3 = 12\,\mbox{.}</math></center></li>
<center><math>\int_{0}^{4} 3 \, dx = 4 \cdot 3 = 12\,\mbox{.}</math></center></li>
</ol>
</ol>
Zeile 163: Zeile 164:
<ol type="a" start=2>
<ol type="a" start=2>
<li><math>\int_{2}^{5} \Bigl(\frac{x}{2} -1 \Bigr) \, dx</math> <br><br>
<li><math>\int_{2}^{5} \Bigl(\frac{x}{2} -1 \Bigr) \, dx</math> <br><br>
-
Integralen kan tolkas som arean under linjen <math>y=x/2-1</math> från
+
The integral can be interpreted as the area below the line <math>y=x/2-1</math> going from
-
<math>x = 2</math> till <math>x = 5</math>,
+
<math>x = 2</math> to <math>x = 5</math>,
-
dvs. en triangel med basen 3 och höjden 1,5 <br>
+
e. a triangle with a base 3 and a height <br>
<center><math>\int_{2}^{5} \Bigl(\frac{x}{2} -1 \Bigr) \, dx
<center><math>\int_{2}^{5} \Bigl(\frac{x}{2} -1 \Bigr) \, dx
= \frac{3 \cdot 1{,}5}{2} = 2{,}25\,\mbox{.}</math></center></li>
= \frac{3 \cdot 1{,}5}{2} = 2{,}25\,\mbox{.}</math></center></li>
Zeile 178: Zeile 179:
<li><math>\int_{0}^{a} kx \, dx\,\mbox{}\quad</math> där
<li><math>\int_{0}^{a} kx \, dx\,\mbox{}\quad</math> där
<math>k>0\,</math>.<br><br>
<math>k>0\,</math>.<br><br>
-
Integralen kan tolkas som arean under linjen <math>y=kx</math> från
+
The integral can be interpreted as the area below the line <math>y=kx</math> going from
-
<math>x = 0</math> till <math>x = a</math>, dvs. en triangel
+
<math>x = 0</math> to <math>x = a</math>, that is a triangle with a base <math>a</math> and a height <math>ka</math><br>
-
med basen <math>a</math> och höjden <math>ka</math><br>
+
<center><math>\int_{0}^{\,a} kx\,dx = \frac{a \cdot ka}{2}
<center><math>\int_{0}^{\,a} kx\,dx = \frac{a \cdot ka}{2}
= \frac{ka^2}{2}\,\mbox{.}</math></center></li>
= \frac{ka^2}{2}\,\mbox{.}</math></center></li>
Zeile 191: Zeile 191:
-
== Primitiv funktion ==
+
== The primitive function ==
-
Funktionen <math>F</math> är en ''primitiv'' funktion till <math>f</math> om <math>F'(x) = f(x)</math> i något intervall. Om <math>F(x)</math> är en primitiv funktion till <math>f(x)</math> så är det klart att även <math>F(x) + C</math> är det, för varje konstant <math>C</math>. Dessutom kan man visa att <math>F(x) + C</math> beskriver samtliga primitiva funktioner till <math>f(x)</math>.
+
The function <math>F</math> is a ''primitive'' function for <math>f</math> if <math>F'(x) = f(x)</math> in any interval. If <math>F(x)</math> is a primitive function for <math>f(x)</math>, it is clear that <math>F(x) + C</math> is as well, for any constant <math>C</math>. In addition, it can be shown that <math>F(x) + C</math> gives all possible primitive functions of <math>f(x)</math>.
<div class="exempel">
<div class="exempel">
'''Exempel 6'''
'''Exempel 6'''
<ol type="a">
<ol type="a">
-
<li><math>F(x) = x^3 + \cos x - 5</math> är en primitiv funktion till
+
<li><math>F(x) = x^3 + \cos x - 5</math> is a primitive function of
-
<math>f(x) = 3x^2 - \sin x</math>, eftersom
+
<math>f(x) = 3x^2 - \sin x</math>, because
{{Fristående formel||<math>F'(x) = D\,(x^3+\cos x-5) = 3x^2-\sin x-0
{{Fristående formel||<math>F'(x) = D\,(x^3+\cos x-5) = 3x^2-\sin x-0
= f(x)\,\mbox{.}</math>}}</li>
= f(x)\,\mbox{.}</math>}}</li>
-
<li><math>G(t) = e^{3t + 1} + \ln t</math> är en primitiv funktion
+
<li><math>G(t) = e^{3t + 1} + \ln t</math> is a primitive function of <math>g(t)= 3 e^{3t + 1} + 1/t</math>, because
-
till <math>g(t)= 3 e^{3t + 1} + 1/t</math>, eftersom
+
{{Fristående formel||<math>G'(t) = D\,\bigl(e^{3t+1}+\ln t\bigr)
{{Fristående formel||<math>G'(t) = D\,\bigl(e^{3t+1}+\ln t\bigr)
= e^{3t+1}\cdot 3+\frac{1}{t} = g(t)\,\mbox{.}</math>}}</li>
= e^{3t+1}\cdot 3+\frac{1}{t} = g(t)\,\mbox{.}</math>}}</li>
-
<li><math>F(x) = \frac{1}{4}x^4 - x + C\,</math>, där <math>C</math> är
+
<li><math>F(x) = \frac{1}{4}x^4 - x + C\,</math>, where <math>C</math> is an arbitrary constant, gives all the primitive functions of <math>f(x) = x^3 - 1</math>.</li>
-
en godtycklig konstant, beskriver samtliga primitiva
+
-
funktioner till <math>f(x) = x^3 - 1</math>.</li>
+
</ol>
</ol>
Zeile 214: Zeile 211:
-
== Samband mellan integral och primitiv funktion ==
+
== The relationship between an integral and its primitive function==
-
Vi har tidigare konstaterat att arean under en funktionskurva, dvs. integralen av en funktion, är beroende av funktionskurvans utseende. Det visar sig att detta beroende utnyttjar den primitiva funktionen, vilket också ger oss möjligheten att beräkna en sådan area exakt.
+
We have previously found that the area below the curve of a function, i.e. integral of a function, is dependent on the form of the curve. It turns out that this dependence makes use of the primitive function, which also gives rise to the ability to calculate such an area exactly.
-
Antag att <math>f</math> är en kontinuerlig funktion på ett intervall (= funktionskurvan har inga avbrott i intervallet). Värdet av integralen <math>\ \int_{a}^{b} f(x) \, dx\ </math> är då beroende av integrationsgränserna <math>a</math> och <math>b</math>, men om man låter <math>a</math> vara ett fixt värde och sätter <math>x</math> som övre gräns blir integralens värde beroende enbart av den övre integrationsgränsen. För att tydliggöra detta använder vi här i stället <math>t</math> som integrationsvariabel:
+
Suppose that <math>f</math> is a continuous function in an interval The value of integral <math>\ \int_{a}^{b} f(x) \, dx\ </math>then is dependent on the limits of integration <math>a</math> and <math>b</math>, but if one lets <math>a</math> have a fixed value and lets <math>x</math> be the upper limit, the integral will depend only on the upper limit. To clarify this, we prefer to use <math>t</math> as the variable of integration:
<center>{{:2.1 - Figur - Area under grafen y = f(x) från t = a till t = x}}</center>
<center>{{:2.1 - Figur - Area under grafen y = f(x) från t = a till t = x}}</center>
Zeile 224: Zeile 221:
{{Fristående formel||<math>A(x) = \int_{a}^{\,x} f(t) \, dt\,\mbox{.}</math>}}
{{Fristående formel||<math>A(x) = \int_{a}^{\,x} f(t) \, dt\,\mbox{.}</math>}}
-
Vi ska nu visa att <math>A</math> i själva verket är en primitiv funktion till <math>f</math>.
+
We shall now show that <math>A</math> is in fact a primitive function of <math>f</math>.
<center>{{:2.1 - Figur - Area under grafen y = f(x) från t = a till t = x + h}}</center>
<center>{{:2.1 - Figur - Area under grafen y = f(x) från t = a till t = x + h}}</center>
-
Den totala arean under kurvan från <math>t=a</math> till <math>t=x+h</math> kan skrivas som <math>A(x+h)</math> och är approximativt lika med arean fram till <math>t=x</math> plus arean av en stapel från <math>t=x</math> till <math>t=x+h</math>, dvs.
+
The total area below the curve from <math>t=a</math> to <math>t=x+h</math> can be written as <math>A(x+h)</math> and is approximately equal to the area up to <math>t=x</math> plus the area of the column between <math>t=x</math> and <math>t=x+h</math>, i.e. .
{{Fristående formel||<math>A(x+h)\approx A(x)+h\cdot f(c)</math>}}
{{Fristående formel||<math>A(x+h)\approx A(x)+h\cdot f(c)</math>}}
-
där <math>c</math> är ett tal mellan <math>x</math> och <math>x+h</math>. Detta uttryck kan vi skriva om som
+
where <math>c</math> is a number between <math>x</math> and <math>x+h</math>. This expression can be rewriten as
{{Fristående formel||<math>\frac{A(x+h)-A(x)}{h} = f(c)\,\mbox{.}</math>}}
{{Fristående formel||<math>\frac{A(x+h)-A(x)}{h} = f(c)\,\mbox{.}</math>}}
-
Om vi låter <math>h \rightarrow 0</math> så går vänstra ledet mot <math>A'(x)</math> och det högra ledet mot <math>f(x)</math> , dvs.
+
If we let <math>h \rightarrow 0</math> the the left-hand side tends towards <math>A'(x)</math> and the right-hand side tends towards <math>f(x)</math> , i.e. .
{{Fristående formel||<math>A'(x) = f(x)\,\mbox{.}</math>}}
{{Fristående formel||<math>A'(x) = f(x)\,\mbox{.}</math>}}
-
Funktionen <math>A(x)</math> är alltså en primitiv funktion till <math>f(x)</math>.
+
Thus the function <math>A(x)</math> is a primitive function of <math>f(x)</math>.
-
== Beräkning av integraler ==
+
== Evaluating integrals ==
-
För att kunna använda primitiva funktioner vid beräkning av en bestämd integral, noterar vi först att om <math>F</math> är en primitiv funktion till <math>f</math> så är
+
In order to use primitive functions in the calculation of a definite integral, we note first that if <math>F</math> is a primitive function of <math>f</math> then
{{Fristående formel||<math>\int_{a}^{\,b} f(t) \, dt = F(b) + C</math>}}
{{Fristående formel||<math>\int_{a}^{\,b} f(t) \, dt = F(b) + C</math>}}
-
där konstanten <math>C</math> måste väljas så att högerledet blir noll när <math>b=a</math>, dvs.
+
where the constant <math>C</math> must be chosen so that the right-hand side is zero when <math>b=a</math>, i.e.
{{Fristående formel||<math>\int_{a}^{\,a} f(t) \, dt = F(a) + C = 0</math>}}
{{Fristående formel||<math>\int_{a}^{\,a} f(t) \, dt = F(a) + C = 0</math>}}
-
vilket ger att <math>C=-F(a)</math>. Om vi sammanfattar har vi alltså att
+
which gives that <math>C=-F(a)</math>. If we summarise, we have that
{{Fristående formel||<math>\int_{a}^{\,b} f(t) \, dt
{{Fristående formel||<math>\int_{a}^{\,b} f(t) \, dt
= F(b) - F(a)\,\mbox{.}</math>}}
= F(b) - F(a)\,\mbox{.}</math>}}
-
Vi kan naturligtvis här lika gärna välja <math>x</math> som integrationsvariabel och skriva
+
We can, of course, just as easily, choose <math>x</math> as the variable of integration and write
{{Fristående formel||<math>\int_{a}^{\,b} f(x) \, dx
{{Fristående formel||<math>\int_{a}^{\,b} f(x) \, dx
= F(b) - F(a)\,\mbox{.}</math>}}
= F(b) - F(a)\,\mbox{.}</math>}}
-
Vid beräkning av integraler utför man detta i två steg. Först bestämmer man en primitiv funktion och sedan sätter man in integrationsgränserna. Man skriver vanligtvis
+
Evaluating an integral is performed in two steps. First one determines a primitive function, and then inserts the limits of integration. The usual way of writing this is as follows,
{{Fristående formel||<math>\int_{a}^{\,b} f(x) \, dx
{{Fristående formel||<math>\int_{a}^{\,b} f(x) \, dx
Zeile 270: Zeile 267:
<div class="exempel">
<div class="exempel">
-
'''Exempel 7'''
+
''' Example 7'''
-
Arean som begränsas av kurvan <math>y=2x - x^2</math> och ''x''-axeln kan beräknas med hjälp av integralen
+
The area bounded by the curve <math>y=2x - x^2</math> and the ''x''-axis can be calculated by using the integral
{| width="100%"
{| width="100%"
| width="95%" |
| width="95%" |
{{Fristående formel||<math>\int_{0}^{2} (2x-x^2) \, dx\,\mbox{.}</math>}}
{{Fristående formel||<math>\int_{0}^{2} (2x-x^2) \, dx\,\mbox{.}</math>}}
-
Eftersom <math>x^2-x^3/3</math> är en primitiv funktion till integranden är integralens värde
+
Since <math>x^2-x^3/3</math> is a primitive function of the integrand the integrals value is
{{Fristående formel||<math>\begin{align*}\int_{0}^{2} (2x-x^2) \, dx &= \Bigl[\,x^2 - {\textstyle\frac{1}{3}}x^3\, \Bigr]_{0}^{2}\\[4pt] &= \bigl( 2^2 - \tfrac{1}{3}2^3\bigr) - \bigl(0^2-\tfrac{1}{3}0^3\bigr)\\[4pt] &= 4 - \tfrac{8}{3} = \tfrac{4}{3}\,\mbox{.}\end{align*}</math>}}
{{Fristående formel||<math>\begin{align*}\int_{0}^{2} (2x-x^2) \, dx &= \Bigl[\,x^2 - {\textstyle\frac{1}{3}}x^3\, \Bigr]_{0}^{2}\\[4pt] &= \bigl( 2^2 - \tfrac{1}{3}2^3\bigr) - \bigl(0^2-\tfrac{1}{3}0^3\bigr)\\[4pt] &= 4 - \tfrac{8}{3} = \tfrac{4}{3}\,\mbox{.}\end{align*}</math>}}
-
Arean är <math>\frac{4}{3}</math>&nbsp;a.e.
+
The area is<math>\frac{4}{3}</math>&nbsp;u.a.
| width="5%" |
| width="5%" |
||{{:2.1 - Figur - Area under grafen y = 2x - x² från x = 0 till x = 2}}
||{{:2.1 - Figur - Area under grafen y = 2x - x² från x = 0 till x = 2}}
|}
|}
-
''Anm:'' Integralvärdet har ingen enhet. I praktiska tillämpningar kan dock arean ha en enhet. Om arean i en enhetslös figur efterfrågas skriver man ofta ''a.e. (areaenheter)'' efter siffervärdet.
+
''Note:'' The value of the integral contains no unit. In practical applications, however, the area may have a unit. If the area in a figure without units is to be obtained one sometimes writes u.a. (units of area) after the value.
-
 
+
</div>
</div>
-
== Baklängesderivering ==
+
== Antidifferentiation ==
-
Att derivera de vanliga funktionstyperna innebär inga oöverstigliga problem; det finns generella metoder för detta. Att utföra den omvända operationen, dvs. hitta en primitiv funktion till en given funktion är dock betydligt svårare och i vissa fall omöjligt! Det finns ingen systematisk metod som fungerar överallt, men genom att utnyttja de vanliga deriveringsreglerna "baklänges" och dessutom lära sig ett antal specialmetoder och knep kan man klara av en stor del av de funktioner som vanligtvis förekommer.
+
To differentiate the common functions is not an insurmountable problems, there are general methods for doing this. To perform the reverse operation, that is find a primitive function for a given function, however, is much more difficult and in some cases impossible! There is no systematic method that works everywhere, but by exploiting the usual rules of differentiation "in the opposite direction" and also by learning a number of special techniques and tricks one can tackle a large number of the functions that turn up.
-
Symbolen <math>\,\int f(x) \,dx\ </math> kallas den ''obestämda'' integralen av <math>f(x)</math> och används för att beteckna en godtycklig primitiv funktion till <math>f(x)</math>. De vanliga deriveringsreglerna ger att
+
The symbol <math>\,\int f(x) \,dx\ </math> is called the ''indefinite'' integral of <math>f(x)</math> and is used to denote an arbitrary primitive function for <math>f(x)</math>. The usual rules of differentiation give
{{Fristående formel||<math>\begin{align*}\int x^n \, dx &= \frac{x^{n+1}}{n+1} + C \quad \text{där }\ n \ne -1\\[6pt] \int x^{-1} \, dx &= \ln |x| + C\\[6pt] \int e^x \, dx &= e^x + C\\[6pt] \int \cos x \, dx &= \sin x + C\\[6pt] \int \sin x \, dx &= -\cos x + C \end{align*}</math>}}
{{Fristående formel||<math>\begin{align*}\int x^n \, dx &= \frac{x^{n+1}}{n+1} + C \quad \text{där }\ n \ne -1\\[6pt] \int x^{-1} \, dx &= \ln |x| + C\\[6pt] \int e^x \, dx &= e^x + C\\[6pt] \int \cos x \, dx &= \sin x + C\\[6pt] \int \sin x \, dx &= -\cos x + C \end{align*}</math>}}
<div class="exempel">
<div class="exempel">
-
'''Exempel 8'''
+
''' Example 8'''
<ol type="a">
<ol type="a">
<li><math>\int (x^4 - 2x^3 + 4x - 7)\,dx
<li><math>\int (x^4 - 2x^3 + 4x - 7)\,dx
Zeile 323: Zeile 319:
-
== Kompensation för ”inre derivata”==
+
== Compensating for the ”inner derivative”==
-
Vid derivering av en sammansatt funktion använder man sig av ''kedjeregeln'', som innebär att man '''multiplicerar''' med den ''inre derivatan''. Om den inre funktionen då är linjär så blir den inre derivatan en konstant. Vid integrering av en sådan funktion måste man därför '''dividera''' med den inre derivatan för att kompensera för detta.
+
When differentiating a composite function one makes use of the ''chain rule'', which means that one must '''multiply''' by the ''inner derivative''. If the inner function is linear, then the inner derivative is a constant. Thus when integrating a composite function, one must '''divide''' by the inner derivative as a sort of compensation.
<div class="exempel">
<div class="exempel">
-
'''Exempel 9'''
+
''' Example 9'''
<ol type="a">
<ol type="a">
Zeile 339: Zeile 335:
<div class="exempel">
<div class="exempel">
-
'''Exempel 10'''
+
''' Example 10'''
<ol type="a">
<ol type="a">
<li><math>\int \sin kx \, dx = - \frac{\cos kx}{k} + C</math></li>
<li><math>\int \sin kx \, dx = - \frac{\cos kx}{k} + C</math></li>
Zeile 348: Zeile 344:
</div>
</div>
-
Observera att detta sätt att kompensera för den inre derivatan endast fungerar om den inre derivatan är en konstant.
+
Note that this way to compensate for the inner derivative only work if the inner derivative is a constant.
-
== Räkneregler för integraler ==
+
== Rules for evaluating integrals ==
-
Med hjälp av beräkningsformeln för integraler är det lätt att visa följande räkneregler för integraler:
+
Using the way integration has been defined here, it is easy to show the following properties of integration:
# <math>\int_{b}^{\,a} f(x) \, dx = - \int_{a}^{\,b} f(x) \, dx\,\mbox{,}\vphantom{\Biggl(}</math>
# <math>\int_{b}^{\,a} f(x) \, dx = - \int_{a}^{\,b} f(x) \, dx\,\mbox{,}\vphantom{\Biggl(}</math>
Zeile 360: Zeile 356:
# <math>\int_{a}^{\,b} f(x) \, dx + \int_{b}^{\,c} f(x)\, dx = \int_{a}^{\,c} f(x)\, dx\,\mbox{.}</math>
# <math>\int_{a}^{\,b} f(x) \, dx + \int_{b}^{\,c} f(x)\, dx = \int_{a}^{\,c} f(x)\, dx\,\mbox{.}</math>
-
Dessutom gäller att area under ''x''-axeln räknas negativt, dvs. om funktionskurvan ligger under ''x''-axeln så blir integralens värde negativt:
+
Moreover, areas below the ''x''- axis are subtracted, that is. if the curve of the function lies below the ''x''-axis in a region, the integral has a negative value in this region:
{| align="center"
{| align="center"
Zeile 368: Zeile 364:
|}
|}
-
Den sammanlagda arean blir <math>\ A_1 + A_2 = \int_{a}^{\,b} f(x)\, dx - \int_{b}^{\,c} f(x)\, dx\,</math>.
+
The total area is <math>\ A_1 + A_2 = \int_{a}^{\,b} f(x)\, dx - \int_{b}^{\,c} f(x)\, dx\,</math>.
-
''Anm.'' Värdet av en '''integral''' kan alltså vara negativt, medan en '''area''' alltid har ett positivt värde.
+
''Note .'' The value of an '''integral''' can be negative, while an '''area''' always has a positive value.
<div class="exempel">
<div class="exempel">
-
'''Exempel 11'''
+
''' Example 11'''
<ol type="a">
<ol type="a">
<li><math>\int_{1}^{2} (x^3 - 3x^2 + 2x + 1) \, dx + \int_{1}^{2} 2 \, dx
<li><math>\int_{1}^{2} (x^3 - 3x^2 + 2x + 1) \, dx + \int_{1}^{2} 2 \, dx
Zeile 387: Zeile 383:
| align="center" |{{:2.1 - Figur - Area för y = x³ - 3x² + 2x + 1, y = 2 och y = x³ - 3x² + 2x + 3}}
| align="center" |{{:2.1 - Figur - Area för y = x³ - 3x² + 2x + 1, y = 2 och y = x³ - 3x² + 2x + 3}}
|-
|-
-
||<small>Den vänstra figuren visar arean under grafen till ''f''(''x'')&nbsp;= ''x''³&nbsp;- 3''x''²&nbsp;+ 2''x''&nbsp;+&nbsp;1 och den mittersta figuren visar arean under grafen till ''g''(''x'')&nbsp;=&nbsp;2. I figuren till höger adderas dessa areor ihop och ger arean under grafen till ''f''(''x'')&nbsp;+&nbsp;''g''(''x'').</small>
+
||<small> The left diagram shows the area below the graph for ''f''(''x'')&nbsp;= ''x''³&nbsp;- 3''x''²&nbsp;+ 2''x''&nbsp;+&nbsp;1 and the middle diagram shows the area below the graph for ''g''(''x'')&nbsp;=&nbsp;2. In the diagram on the right these areas are summed and give the area below the graph for ''f''(''x'')&nbsp;+&nbsp;''g''(''x'').</small>
|}
|}
</li>
</li>
Zeile 401: Zeile 397:
| align="center" |{{:2.1 - Figur - Area för y = x²/2 - 2x, y = 2x - x²/2 + 3/2 och y = 3/2}}
| align="center" |{{:2.1 - Figur - Area för y = x²/2 - 2x, y = 2x - x²/2 + 3/2 och y = 3/2}}
|-
|-
-
||<small>Grafen till ''f''(''x'')&nbsp;= ''x''²/2&nbsp;- 2''x'' (figuren till vänster) och grafen till ''g''(''x'')&nbsp;= 2''x''&nbsp;- ''x''²/2&nbsp;+ 3/2 (figuren i mitten) är spegelsymmetriska kring linjen ''y''&nbsp;= 3/4 (streckad linje i figurerna) och det gör att summan ''f''(''x'')&nbsp;+ ''g''(''x'') är konstant lika med&nbsp;3/2. Summan av integralernas värde är därför lika med arean av en rektangel med bas&nbsp;2 och höjd&nbsp;3/2 (figuren till höger).</small>
+
||<small>The graph to ''f''(''x'')&nbsp;= ''x''²/2&nbsp;- 2''x'' (diagram on the left) and the graph to ''g''(''x'')&nbsp;= 2''x''&nbsp;- ''x''²/2&nbsp;+ 3/2 ( diagram in the middle) are inverted with respect to each other about the line ''y''&nbsp;= 3/4 (dotted line in the diagrams). This means the sum ''f''(''x'')&nbsp;+ ''g''(''x'') is equal to&nbsp;3/2. and is a constant. Thus the sum of the integrals is equal to the area of a rectangle with base &nbsp;2 and height&nbsp;3/2 ( diagram on the right). </small>
|}
|}
Zeile 430: Zeile 426:
| align="center" |{{:2.1 - Figur - Area för y = x² - 1}}
| align="center" |{{:2.1 - Figur - Area för y = x² - 1}}
|-
|-
-
||<small>Figuren visar grafen till f(x) = x² - 1 och beräkningen ovan visar att den skuggade arean under ''x''-axeln är lika stor som den skuggade arean ovanför ''x''-axeln.</small>
+
||<small> The figure shows the graph of f(x) = x² - 1 and the calculation above shows that the shaded area below the ''x''-axis is equal to the shaded area above the ''x''-axis.</small>
|}
|}
</li>
</li>
Zeile 438: Zeile 434:
-
== Area mellan kurvor ==
+
== Area between curves ==
-
Om <math>f(x) \ge g(x)</math> i ett intervall <math>a\le x\le b</math> gäller att arean mellan funktionskurvorna ges av
+
If <math>f(x) \ge g(x)</math> in an interval <math>a\le x\le b</math> then that the area between the curves of the function is given by
{{Fristående formel||<math>\int_{a}^{b} f(x) \, dx
{{Fristående formel||<math>\int_{a}^{b} f(x) \, dx
- \int_{a}^{b} g(x) \, dx\,\mbox{,}</math>}}
- \int_{a}^{b} g(x) \, dx\,\mbox{,}</math>}}
-
vilket kan förenklas till
+
which can be simplified to
{{Fristående formel||<math>\int_{a}^{b} (f(x) - g(x)) \, dx\,\mbox{.}</math>}}
{{Fristående formel||<math>\int_{a}^{b} (f(x) - g(x)) \, dx\,\mbox{.}</math>}}
Zeile 451: Zeile 447:
<center>{{:2.1 - Figur - Area mellan y = f(x) och y = g(x)}}</center>
<center>{{:2.1 - Figur - Area mellan y = f(x) och y = g(x)}}</center>
{| width="90%" align="center"
{| width="90%" align="center"
-
||<small>Om ''f''(''x'') och ''g''(''x'') antar positiva värden och ''f''(''x'') är större än ''g''(''x''), då ges arean mellan graferna till ''f'' och ''g'' (figuren till vänster) som differensen mellan arean under grafen till ''f'' (figuren i mitten) och arean under grafen till ''g'' (figuren till höger).</small>
+
||<small>If ''f''(''x'') and ''g''(''x'') take positive values and ''f''(''x'') is greater than ''g''(''x''), the area between the graphs of ''f'' and ''g'' (the figure on the left) can be obtained as the difference between the area below the graph ''f'' (figure in the middle) and the area below the graph ''g'' (the figure on the right).</small>
|}
|}
-
Observera att det inte spelar någon roll om <math>f(x) < 0</math> eller <math>g(x) < 0</math> så länge som <math>f(x) \ge g(x)</math>. Arean mellan kurvorna är naturligtvis lika stor oavsett om kurvorna ligger över eller under ''x''-axeln, vilket följande figurer illustrerar:
+
Note that it does not matter whether <math>f(x) < 0</math> or <math>g(x) < 0</math> as long as <math>f(x) \ge g(x)</math>. The value of the area between the curves is independent of whether the curves are above or below the ''x''-xis, as the following figures illustrate:
<center>{{:2.1 - Figur - Area translaterad i y-led}}</center>
<center>{{:2.1 - Figur - Area translaterad i y-led}}</center>
{| width="90%" align="center"
{| width="90%" align="center"
-
||<small>Arean mellan två grafer påverkas inte av om graferna translateras i ''y''-led. Arean mellan graferna till f(x) och g(x) (figuren till vänster) är lika med arean mellan graferna till f(x) - 3 och g(x) - 3 (figuren i mitten), likväl som arean mellan graferna till f(x) - 6 och g(x) - 6 (figuren till höger).</small>
+
||<small>The area between the two graphs is not affected if the graphs are moved in the ''y''-direction. The area between the graphs of f(x) and g(x) (figure on the left) is equal to the area between the graphs of f(x) - 3 and g(x) - 3 (the figure in the middle), as well as the area between the graphs of f(x) - 6 and g(x) - 6 (figure on the right).</small>
|}
|}
<div class="exempel">
<div class="exempel">
-
'''Exempel 12'''
+
''' Example 12'''
-
Beräkna arean av det område som begränsas av kurvorna <math>y=e^x + 1</math> och <math>y=1 - x^2/2</math> samt linjerna <math>x = –1</math> och <math>x = 1</math>.
+
Calculate the area bounded by the curves <math>y=e^x + 1</math> and <math>y=1 - x^2/2</math> and the lines <math>x = –1</math> and <math>x = 1</math>.
<br>
<br>
<br>
<br>
-
Eftersom <math>e^x + 1 > 1 - x^2/2</math> i hela intervallet blir områdets area
+
Since <math>e^x + 1 > 1 - x^2/2</math> in all the interval the area in question is given by
{| width="100%"
{| width="100%"
| width="95%" |
| width="95%" |
Zeile 479: Zeile 475:
<div class="exempel">
<div class="exempel">
-
'''Exempel 13'''
+
''' Example 13'''
-
Beräkna arean av det ändliga område som begränsas av kurvorna <math>y= x^2</math> och <math>y= \sqrt[\scriptstyle 3]{x}</math>.
+
Calculate the area of the finite area bounded by the curves <math>y= x^2</math> and <math>y= \sqrt[\scriptstyle 3]{x}</math>.
<br>
<br>
<br>
<br>
-
Kurvorna skär varandra i punkter där deras ''y''-värden är lika
+
The curves intersect at the points where their ''y''-values are equal
-
{{Fristående formel||<math>\begin{align*} &x^2 = x^{1/3} \quad \Leftrightarrow \quad x^6 = x\quad \Leftrightarrow \quad x(x^5 - 1) = 0\\ &\quad \Leftrightarrow \quad x=0 \quad \text{eller}\quad x=1\,\mbox{.}\end{align*}</math>}}
+
{{Fristående formel||<math>\begin{align*} &x^2 = x^{1/3} \quad \Leftrightarrow \quad x^6 = x\quad \Leftrightarrow \quad x(x^5 - 1) = 0\\ &\quad \Leftrightarrow \quad x=0 \quad \text{or}\quad x=1\,\mbox{.}\end{align*}</math>}}
{| width="100%"
{| width="100%"
-
| width="95%" |Mellan <math>x=0</math> och <math>x=1</math> är <math>\sqrt[\scriptstyle 3]{x}>x^2</math> så områdets area ges av
+
| width="95%" | Between <math>x=0</math> and <math>x=1</math>, <math>\sqrt[\scriptstyle 3]{x}>x^2</math> is true, thus the area is
{{Fristående formel||<math>\begin{align*}\int_{0}^{1} \bigl( x^{1/3} - x^2 \bigr) \, dx &= \Bigl[\,\frac{ x^{4/3}}{4/3} - \frac{x^3}{3}\,\Bigr]_{0}^{1}\\
{{Fristående formel||<math>\begin{align*}\int_{0}^{1} \bigl( x^{1/3} - x^2 \bigr) \, dx &= \Bigl[\,\frac{ x^{4/3}}{4/3} - \frac{x^3}{3}\,\Bigr]_{0}^{1}\\
Zeile 503: Zeile 499:
<div class="exempel">
<div class="exempel">
-
'''Exempel 14'''
+
''' Example 14'''
-
Beräkna arean av det område som begränsas av kurvan <math>y=\frac{1}{x^2}</math> samt linjerna <math>y=x</math> och <math>y = 2</math>.
+
Calculate the area of the region bounded by the curve <math>y=\frac{1}{x^2}</math>and the lines <math>y=x</math> and <math>y = 2</math>.
<br>
<br>
<br>
<br>
{| width="100%"
{| width="100%"
| width="95%" |
| width="95%" |
-
I figuren till höger är kurvan och de två linjerna skisserade och då ser vi att området kan delas upp i två delområden som var och en ligger mellan två funktionskurvor. Den totala arean är därför summan av integralerna
+
 
 +
In the figure on the right, the curve and the two lines have been sketched and then we see that the region can be divided into two sub-regions, each of which is located between two curves. The total area is the sum of the integrals
{{Fristående formel||<math>A_1 = \int_{a}^{\,b} (2 - \frac{1}{x^2}) \, dx
{{Fristående formel||<math>A_1 = \int_{a}^{\,b} (2 - \frac{1}{x^2}) \, dx
\quad\text{och}\quad A_2 = \int_{b}^{\,c} (2- x) \, dx\,\mbox{.}</math>}}
\quad\text{och}\quad A_2 = \int_{b}^{\,c} (2- x) \, dx\,\mbox{.}</math>}}
-
Vi bestämmer först skärningspunkterna <math>x=a</math>, <math>x=b</math> och <math>x=c</math>:
+
We first determine the points of intersection <math>x=a</math>, <math>x=b</math> and <math>x=c</math>:
| width="5%" |
| width="5%" |
||{{:2.1 - Figur - Area som begränsas av y = 1/x², y = x och y = 2}}
||{{:2.1 - Figur - Area som begränsas av y = 1/x², y = x och y = 2}}
|}
|}
-
*Skärningspunkten <math>x=a</math> bestäms av ekvationen
+
* The point of intersection <math>x=a</math> is obtained from the equation
{{Fristående formel||<math>\frac{1}{x^2} = 2
{{Fristående formel||<math>\frac{1}{x^2} = 2
Zeile 525: Zeile 522:
\quad \Leftrightarrow \quad x = \pm \frac{1}{\sqrt{2}}\,\mbox{.}</math>}}
\quad \Leftrightarrow \quad x = \pm \frac{1}{\sqrt{2}}\,\mbox{.}</math>}}
-
:(Den negativa roten är dock inte aktuell.)
+
:(The negative root, however, is not relevant.)
-
*Skärningspunkt <math>x=b</math> bestäms av ekvationen
+
* The point of intersection <math>x=b</math> is obtained from the equation
{{Fristående formel||<math>\frac{1}{x^2} = x
{{Fristående formel||<math>\frac{1}{x^2} = x
Zeile 533: Zeile 530:
\quad \Leftrightarrow \quad x=1\,\mbox{.}</math>}}
\quad \Leftrightarrow \quad x=1\,\mbox{.}</math>}}
-
*Skärningspunkt <math>x=c</math> bestäms av ekvationen <math>x = 2</math>.
+
*The point of intersection <math>x=c</math> is obtained from the equation <math>x = 2</math>.
-
Integralerna blir därför
+
The integrals are therefore
{{Fristående formel||<math>\begin{align*} A_1 &= \int_{1/\sqrt{2}}^{1} \Bigl(2 - \frac{1}{x^2}\Bigr) \, dx = \int_{1/\sqrt{2}}^{1} \bigl(2 - x ^{-2}\bigr) \, dx = \Bigl[\,2x-\frac{x^{-1}}{-1}\,\Bigr]_{1/\sqrt{2}}^{1}\\[4pt] &= \Bigl[\,2x + \frac{1}{x}\,\Bigr]_{1/\sqrt{2}}^{1} = (2+ 1) - \Bigl( \frac{2}{\sqrt{2}} + \sqrt{2}\,\Bigr) = 3 - 2\sqrt{2}\,\mbox{,}\\[4pt] A_2 &= \int_{1}^{2} (2 - x) \, dx = \Bigl[\,2x - \frac{x^2}{2}\,\Bigr]_{1}^{2} = (4-2) - \Bigl(2- \frac{1}{2}\Bigr) = \frac{1}{2}\,\mbox{.}
{{Fristående formel||<math>\begin{align*} A_1 &= \int_{1/\sqrt{2}}^{1} \Bigl(2 - \frac{1}{x^2}\Bigr) \, dx = \int_{1/\sqrt{2}}^{1} \bigl(2 - x ^{-2}\bigr) \, dx = \Bigl[\,2x-\frac{x^{-1}}{-1}\,\Bigr]_{1/\sqrt{2}}^{1}\\[4pt] &= \Bigl[\,2x + \frac{1}{x}\,\Bigr]_{1/\sqrt{2}}^{1} = (2+ 1) - \Bigl( \frac{2}{\sqrt{2}} + \sqrt{2}\,\Bigr) = 3 - 2\sqrt{2}\,\mbox{,}\\[4pt] A_2 &= \int_{1}^{2} (2 - x) \, dx = \Bigl[\,2x - \frac{x^2}{2}\,\Bigr]_{1}^{2} = (4-2) - \Bigl(2- \frac{1}{2}\Bigr) = \frac{1}{2}\,\mbox{.}
\end{align*}</math>}}
\end{align*}</math>}}
-
Den sammanlagda arean blir
+
The total area is
{{Fristående formel||<math> A_1 + A_2 = 3 - 2\sqrt{2} + \tfrac{1}{2} = \tfrac{7}{2} - 2\sqrt{2}\ \text{a.e.}</math>}}
{{Fristående formel||<math> A_1 + A_2 = 3 - 2\sqrt{2} + \tfrac{1}{2} = \tfrac{7}{2} - 2\sqrt{2}\ \text{a.e.}</math>}}
</div>
</div>

Version vom 17:23, 21. Jul. 2008

 
  1. REDIRECT Template:Gewählter Tab
  2. REDIRECT Template:Nicht gewählter Tab
 

Content:

  • Definition of an integral (overview).
  • Fundamental theorm of integral calculus.
  • Primitive function for \displaystyle x^\alpha, \displaystyle 1/x, \displaystyle e^x, \displaystyle \cos x and \displaystyle \sin x.
  • Primitive function for sum and difference.

Learning outcomes:

After this section, you will have learned to :

  • Interpret integrals as signed areas, that is, " the area above the \displaystyle x-axis" minus " area below the \displaystyle x-axis".
  • Understand other interpretations of the integral, for example. density / mass, speed / displacement, power / charge , etc.
  • Determine primitive function \displaystyle x^\alpha, \displaystyle 1/x, \displaystyle e^{kx}, \displaystyle \cos kx, \displaystyle \sin kx and the sum / difference of such terms.
  • Calculate the area below the curve of a function.
  • Calculate the area between two curves of two functions.
  • Recognise that all functions do not have primitive functions that can be written as a closed analytical expression, such as \displaystyle e^{x^2} , \displaystyle (\sin x)/x, \displaystyle \sin \sin x, etc.

Area below the curve of a function

We previously have found that the slope of a curve of a function is interesting. It gives us information about how the function changes and has great significance in many applications. In a similar way the area between the curve of a function and the x-axis is of importance. It of course is dependent on the curves appearance and thus closely related to the function in question. It is easy to see that this area has practical significance in many different contexts.

If an object is moving, we can illustrate its speed v plotted against time t in a v,t-diagram. We can see in the figure below three different hypothetical examples:


2.1 - Figur - v-t-diagram med konstant fart 5 2.1 - Figur - v-t-diagram med konstant fart 4 och 6 2.1 - Figur - v-t-diagram med fart v(t) = t
The object moves at a constant speed of 5. The object moves at a steady speed of 4 when an impact at t = 3 suddenly increases the speed to 6. The object is sliding down a sloping plane and has a linearly increasing speed.


The distance travelled is in each case is

  1. REDIRECT Template:Abgesetzte Formel

In each cases, you see that the distance travelled by the object is matched by the area below the curve.

More examples of what the area below a curve can symbolise are shown below.

Example 1


2.1 - Figur - Effekt-tid-diagram 2.1 - Figur - Kraft-sträcka-diagram 2.1 - Figur - Ström-tid-diagram
A solar cell which has been exposed to light of power p will have received energy that is proportional to the area under the above graph. The force F applied to an object along the direction of its motion does work that is proportional to the area under the above graph. A capacitor that is charged by a current i will receive a charge which is proportional to the area under the above graph.


The notation for an integral.

In order to describe the area below the curve of a function in symbolic form one introduces the integral sign \displaystyle \,\smallint\, :

The integral of a positive function \displaystyle f(x) from \displaystyle a to \displaystyle b is understood to mean the area between the curve \displaystyle y=f(x) and the interval of the x-axis between \displaystyle x=a and \displaystyle x=b , and is written with the notation

  1. REDIRECT Template:Abgesetzte Formel

The numbers \displaystyle a and \displaystyle b are called the lower and upper limits of integration respectively, \displaystyle f(x) is called the integrand and \displaystyle x the variable of integration.

Example 2

The area below the curve \displaystyle y=f(x) from \displaystyle x=a to \displaystyle x=c is equal to the area from \displaystyle x=a to \displaystyle x=b plus the area from \displaystyle x=b to \displaystyle x=c. This means that
  1. REDIRECT Template:Abgesetzte Formel
2.1 - Figur - Area under grafen y = f(x) från a till b och c

Example 3

For an object, whose speed is changing according to the function \displaystyle v(t) the distance travelled after 10 s is characterised by the integral
  1. REDIRECT Template:Abgesetzte Formel

Note . We assume that speed and distance are measured using the same units of length.

2.1 - Figur - Area s(10) i ett v-t-diagram

Example 4

Water is flowing into a tank at a rate of \displaystyle f(t) liter/s at the time \displaystyle t. The integral

  1. REDIRECT Template:Abgesetzte Formel

specifies the amount in litres which flows into the tank during the tenth second.

Example 5

Calculate the integrals

  1. \displaystyle \int_{0}^{4} 3 \, dx

    The integral can be interpreted as area below the curve (the line) \displaystyle y=3 going from \displaystyle x = 0 to \displaystyle x = 4, i.e. a rectangle with the base 4 and height 3,
    \displaystyle \int_{0}^{4} 3 \, dx = 4 \cdot 3 = 12\,\mbox{.}
2.1 - Figur - Area under grafen y = 3 från x = 0 till x = 4
  1. \displaystyle \int_{2}^{5} \Bigl(\frac{x}{2} -1 \Bigr) \, dx

    The integral can be interpreted as the area below the line \displaystyle y=x/2-1 going from \displaystyle x = 2 to \displaystyle x = 5, e. a triangle with a base 3 and a height
    \displaystyle \int_{2}^{5} \Bigl(\frac{x}{2} -1 \Bigr) \, dx = \frac{3 \cdot 1{,}5}{2} = 2{,}25\,\mbox{.}
2.1 - Figur - Area under grafen y = x/2 - 1 från x = 2 till x = 5
  1. \displaystyle \int_{0}^{a} kx \, dx\,\mbox{}\quad där \displaystyle k>0\,.

    The integral can be interpreted as the area below the line \displaystyle y=kx going from \displaystyle x = 0 to \displaystyle x = a, that is a triangle with a base \displaystyle a and a height \displaystyle ka
    \displaystyle \int_{0}^{\,a} kx\,dx = \frac{a \cdot ka}{2} = \frac{ka^2}{2}\,\mbox{.}
2.1 - Figur - Area under grafen y = kx från x = 0 till x = a


The primitive function

The function \displaystyle F is a primitive function for \displaystyle f if \displaystyle F'(x) = f(x) in any interval. If \displaystyle F(x) is a primitive function for \displaystyle f(x), it is clear that \displaystyle F(x) + C is as well, for any constant \displaystyle C. In addition, it can be shown that \displaystyle F(x) + C gives all possible primitive functions of \displaystyle f(x).

Exempel 6

  1. \displaystyle F(x) = x^3 + \cos x - 5 is a primitive function of \displaystyle f(x) = 3x^2 - \sin x, because
    1. REDIRECT Template:Abgesetzte Formel
  2. \displaystyle G(t) = e^{3t + 1} + \ln t is a primitive function of \displaystyle g(t)= 3 e^{3t + 1} + 1/t, because
    1. REDIRECT Template:Abgesetzte Formel
  3. \displaystyle F(x) = \frac{1}{4}x^4 - x + C\,, where \displaystyle C is an arbitrary constant, gives all the primitive functions of \displaystyle f(x) = x^3 - 1.


The relationship between an integral and its primitive function

We have previously found that the area below the curve of a function, i.e. integral of a function, is dependent on the form of the curve. It turns out that this dependence makes use of the primitive function, which also gives rise to the ability to calculate such an area exactly.

Suppose that \displaystyle f is a continuous function in an interval The value of integral \displaystyle \ \int_{a}^{b} f(x) \, dx\ then is dependent on the limits of integration \displaystyle a and \displaystyle b, but if one lets \displaystyle a have a fixed value and lets \displaystyle x be the upper limit, the integral will depend only on the upper limit. To clarify this, we prefer to use \displaystyle t as the variable of integration:

2.1 - Figur - Area under grafen y = f(x) från t = a till t = x
  1. REDIRECT Template:Abgesetzte Formel

We shall now show that \displaystyle A is in fact a primitive function of \displaystyle f.

2.1 - Figur - Area under grafen y = f(x) från t = a till t = x + h

The total area below the curve from \displaystyle t=a to \displaystyle t=x+h can be written as \displaystyle A(x+h) and is approximately equal to the area up to \displaystyle t=x plus the area of the column between \displaystyle t=x and \displaystyle t=x+h, i.e. .

  1. REDIRECT Template:Abgesetzte Formel

where \displaystyle c is a number between \displaystyle x and \displaystyle x+h. This expression can be rewriten as

  1. REDIRECT Template:Abgesetzte Formel

If we let \displaystyle h \rightarrow 0 the the left-hand side tends towards \displaystyle A'(x) and the right-hand side tends towards \displaystyle f(x) , i.e. .

  1. REDIRECT Template:Abgesetzte Formel

Thus the function \displaystyle A(x) is a primitive function of \displaystyle f(x).


Evaluating integrals

In order to use primitive functions in the calculation of a definite integral, we note first that if \displaystyle F is a primitive function of \displaystyle f then

  1. REDIRECT Template:Abgesetzte Formel

where the constant \displaystyle C must be chosen so that the right-hand side is zero when \displaystyle b=a, i.e.

  1. REDIRECT Template:Abgesetzte Formel

which gives that \displaystyle C=-F(a). If we summarise, we have that

  1. REDIRECT Template:Abgesetzte Formel

We can, of course, just as easily, choose \displaystyle x as the variable of integration and write

  1. REDIRECT Template:Abgesetzte Formel

Evaluating an integral is performed in two steps. First one determines a primitive function, and then inserts the limits of integration. The usual way of writing this is as follows,

  1. REDIRECT Template:Abgesetzte Formel


Example 7

The area bounded by the curve \displaystyle y=2x - x^2 and the x-axis can be calculated by using the integral

  1. REDIRECT Template:Abgesetzte Formel

Since \displaystyle x^2-x^3/3 is a primitive function of the integrand the integrals value is

  1. REDIRECT Template:Abgesetzte Formel

The area is\displaystyle \frac{4}{3} u.a.

2.1 - Figur - Area under grafen y = 2x - x² från x = 0 till x = 2

Note: The value of the integral contains no unit. In practical applications, however, the area may have a unit. If the area in a figure without units is to be obtained one sometimes writes u.a. (units of area) after the value.


Antidifferentiation

To differentiate the common functions is not an insurmountable problems, there are general methods for doing this. To perform the reverse operation, that is find a primitive function for a given function, however, is much more difficult and in some cases impossible! There is no systematic method that works everywhere, but by exploiting the usual rules of differentiation "in the opposite direction" and also by learning a number of special techniques and tricks one can tackle a large number of the functions that turn up.

The symbol \displaystyle \,\int f(x) \,dx\ is called the indefinite integral of \displaystyle f(x) and is used to denote an arbitrary primitive function for \displaystyle f(x). The usual rules of differentiation give

  1. REDIRECT Template:Abgesetzte Formel

Example 8

  1. \displaystyle \int (x^4 - 2x^3 + 4x - 7)\,dx = \frac{x^5}{5} - \frac{2x^4}{4} + \frac{4x^2}{2} - 7x + C
    \displaystyle \phantom{\int (x^4 - 2x^3 + 4x - 7)\,dx}{} = \frac{x^5}{5} - \frac{x^4}{2} + 2x^2 - 7x + C
  2. \displaystyle \int \Bigl(\frac{3}{x^2} -\frac{1}{2x^3} \Bigr) dx = \int \Bigl( 3x^{-2} - \frac{1}{2} x^{-3} \Bigr) dx = \frac{3x^{-1}}{-1} - \frac{1}{2} \cdot \frac{x^{-2}}{(-2)} + C
    \displaystyle \phantom{\int \Bigl(\frac{3}{x^2} -\frac{1}{2x^3} \Bigr) dx}{} = - 3x^{-1} + \tfrac{1}{4}x^{-2} + C = -\frac{3}{x} + \frac{1}{4x^2} + C\vphantom{\Biggl(}
  3. \displaystyle \int \frac{2}{3x} \,dx = \int \frac{2}{3} \cdot \frac{1}{x} \, dx = \tfrac{2}{3} \ln |x| + C
  4. \displaystyle \int ( e^x - \cos x - \sin x ) \, dx = e^x - \sin x + \cos x +C


Compensating for the ”inner derivative”

When differentiating a composite function one makes use of the chain rule, which means that one must multiply by the inner derivative. If the inner function is linear, then the inner derivative is a constant. Thus when integrating a composite function, one must divide by the inner derivative as a sort of compensation.

Example 9

  1. \displaystyle \int e^{3x} \, dx = \frac{e^{3x}}{3} + C
  2. \displaystyle \int \sin 5x \, dx = - \frac{ \cos 5x}{5} + C
  3. \displaystyle \int (2x +1)^4 \, dx = \frac{(2x+1)^5}{5 \cdot 2} + C

Example 10

  1. \displaystyle \int \sin kx \, dx = - \frac{\cos kx}{k} + C
  2. \displaystyle \int \cos kx \, dx = \frac{\sin kx }{k} + C
  3. \displaystyle \int e^{kx} \, dx = \displaystyle \frac{e^{kx}}{k} + C

Note that this way to compensate for the inner derivative only work if the inner derivative is a constant.


Rules for evaluating integrals

Using the way integration has been defined here, it is easy to show the following properties of integration:

  1. \displaystyle \int_{b}^{\,a} f(x) \, dx = - \int_{a}^{\,b} f(x) \, dx\,\mbox{,}\vphantom{\Biggl(}
  2. \displaystyle \int_{a}^{\,b} f(x) \, dx + \int_{a}^{\,b} g(x) \, dx = \int_{a}^{\,b} (f(x) + g(x)) \, dx\,\mbox{,}\vphantom{\Biggl(}
  3. \displaystyle \int_{a}^{\,b} k \cdot f(x)\, dx = k \int_{a}^{\,b} f(x)\, dx\,\mbox{,}\vphantom{\Biggl(}
  4. \displaystyle \int_{a}^{\,b} f(x) \, dx + \int_{b}^{\,c} f(x)\, dx = \int_{a}^{\,c} f(x)\, dx\,\mbox{.}

Moreover, areas below the x- axis are subtracted, that is. if the curve of the function lies below the x-axis in a region, the integral has a negative value in this region:

\displaystyle \begin{align*}A_1 &= \int_{a}^{\,b} f(x)\, dx,\\[6pt] A_2 &= -\int_{b}^{\,c} f(x)\, dx\,\mbox{.} \end{align*} 2.1 - Figur - Areor A₁ och A₂ mellan y = f(x) och x-axeln

The total area is \displaystyle \ A_1 + A_2 = \int_{a}^{\,b} f(x)\, dx - \int_{b}^{\,c} f(x)\, dx\,.

Note . The value of an integral can be negative, while an area always has a positive value.


Example 11

  1. \displaystyle \int_{1}^{2} (x^3 - 3x^2 + 2x + 1) \, dx + \int_{1}^{2} 2 \, dx =\int_{1}^{2} (x^3 - 3x^2 + 2x + 1+2) \, dx
    \displaystyle \qquad{}= \Bigl[\,\tfrac{1}{4}x^4 - x^3 + x^2 + 3x\,\Bigr]_{1}^{2} \vphantom{\Biggr)^2}
    \displaystyle \qquad{}= \bigl(\tfrac{1}{4}\cdot 4-2^3+2^2+3\cdot 2\bigr) - \bigl(\tfrac{1}{4}\cdot 1^4 - 1^3 + 1^2 + 3\cdot 1\bigr)\vphantom{\Biggr)^2}
    \displaystyle \qquad{}=6-3-\tfrac{1}{4} = \tfrac{11}{4}
    2.1 - Figur - Area för y = x³ - 3x² + 2x + 1, y = 2 och y = x³ - 3x² + 2x + 3
    The left diagram shows the area below the graph for f(x) = x³ - 3x² + 2x + 1 and the middle diagram shows the area below the graph for g(x) = 2. In the diagram on the right these areas are summed and give the area below the graph for f(x) + g(x).


  1. \displaystyle \int_{1}^{3} (x^2/2 - 2x) \, dx + \int_{1}^{3} (2x - x^2/2 + 3/2) \, dx = \int_{1}^{3} 3/2 \, dx
    \displaystyle \qquad{} = \Bigl[\,\tfrac{3}{2}x\,\Bigr]_{1}^{3} = \tfrac{3}{2}\cdot 3 - \tfrac{3}{2}\cdot 1 = 3
    2.1 - Figur - Area för y = x²/2 - 2x, y = 2x - x²/2 + 3/2 och y = 3/2
    The graph to f(x) = x²/2 - 2x (diagram on the left) and the graph to g(x) = 2x - x²/2 + 3/2 ( diagram in the middle) are inverted with respect to each other about the line y = 3/4 (dotted line in the diagrams). This means the sum f(x) + g(x) is equal to 3/2. and is a constant. Thus the sum of the integrals is equal to the area of a rectangle with base  2 and height 3/2 ( diagram on the right).


  1. \displaystyle \int_{1}^{2} \frac{4x^2 - 2}{3x} \, dx = \int_{1}^{2} \frac{2(2x^2-1)}{3x} \, dx = \frac{2}{3} \int_{1}^{2} \frac{2x^2 - 1}{x} \, dx \vphantom{\Biggl(}
    \displaystyle \qquad{}= \frac{2}{3} \int_{1}^{2} \Bigl(2x - \frac{1}{x}\Bigr) \, dx = \frac{2}{3} \Bigl[\,x^2 - \ln x\,\Bigr]_{1}^{2} \vphantom{\Biggl(}
    \displaystyle \qquad{}= \frac {2}{3}\Bigl((4- \ln 2) - (1 - \ln 1)\Bigr) = \tfrac{2}{3}(3 - \ln 2) = 2 - \tfrac{2}{3}\ln 2


  1. \displaystyle \int_{-1}^{2} (x^2 - 1) \, dx = \Bigl[\,\frac{x^3}{3} - x\,\Bigl]_{-1}^{2} = \bigl(\tfrac{8}{3} - 2\bigr) - \bigl(\tfrac{-1}{3} + 1 \bigr) = 0
    2.1 - Figur - Area för y = x² - 1
    The figure shows the graph of f(x) = x² - 1 and the calculation above shows that the shaded area below the x-axis is equal to the shaded area above the x-axis.


Area between curves

If \displaystyle f(x) \ge g(x) in an interval \displaystyle a\le x\le b then that the area between the curves of the function is given by

  1. REDIRECT Template:Abgesetzte Formel

which can be simplified to

  1. REDIRECT Template:Abgesetzte Formel
2.1 - Figur - Area mellan y = f(x) och y = g(x)
If f(x) and g(x) take positive values and f(x) is greater than g(x), the area between the graphs of f and g (the figure on the left) can be obtained as the difference between the area below the graph f (figure in the middle) and the area below the graph g (the figure on the right).


Note that it does not matter whether \displaystyle f(x) < 0 or \displaystyle g(x) < 0 as long as \displaystyle f(x) \ge g(x). The value of the area between the curves is independent of whether the curves are above or below the x-xis, as the following figures illustrate:

2.1 - Figur - Area translaterad i y-led
The area between the two graphs is not affected if the graphs are moved in the y-direction. The area between the graphs of f(x) and g(x) (figure on the left) is equal to the area between the graphs of f(x) - 3 and g(x) - 3 (the figure in the middle), as well as the area between the graphs of f(x) - 6 and g(x) - 6 (figure on the right).

Example 12

Calculate the area bounded by the curves \displaystyle y=e^x + 1 and \displaystyle y=1 - x^2/2 and the lines \displaystyle x = –1 and \displaystyle x = 1.

Since \displaystyle e^x + 1 > 1 - x^2/2 in all the interval the area in question is given by

  1. REDIRECT Template:Abgesetzte Formel
2.1 - Figur - Area mellan y = e^x - 1 och y = 1 - x²/2

Example 13

Calculate the area of the finite area bounded by the curves \displaystyle y= x^2 and \displaystyle y= \sqrt[\scriptstyle 3]{x}.

The curves intersect at the points where their y-values are equal

  1. REDIRECT Template:Abgesetzte Formel
Between \displaystyle x=0 and \displaystyle x=1, \displaystyle \sqrt[\scriptstyle 3]{x}>x^2 is true, thus the area is
  1. REDIRECT Template:Abgesetzte Formel
2.1 - Figur - Area mellan y = ∛x och y = x²

Example 14

Calculate the area of the region bounded by the curve \displaystyle y=\frac{1}{x^2}and the lines \displaystyle y=x and \displaystyle y = 2.

In the figure on the right, the curve and the two lines have been sketched and then we see that the region can be divided into two sub-regions, each of which is located between two curves. The total area is the sum of the integrals

  1. REDIRECT Template:Abgesetzte Formel

We first determine the points of intersection \displaystyle x=a, \displaystyle x=b and \displaystyle x=c:

2.1 - Figur - Area som begränsas av y = 1/x², y = x och y = 2
  • The point of intersection \displaystyle x=a is obtained from the equation
  1. REDIRECT Template:Abgesetzte Formel
(The negative root, however, is not relevant.)
  • The point of intersection \displaystyle x=b is obtained from the equation
  1. REDIRECT Template:Abgesetzte Formel
  • The point of intersection \displaystyle x=c is obtained from the equation \displaystyle x = 2.

The integrals are therefore

  1. REDIRECT Template:Abgesetzte Formel

The total area is

  1. REDIRECT Template:Abgesetzte Formel