From Förberedande kurs i matematik 2

The action you have requested is limited to users in one of the groups Sysops, approved.

You can view and copy the source of this page:

__NOTOC__ {| border="0" cellspacing="0" cellpadding="0" height="30" width="100%" | style="border-bottom:1px solid #797979" width="5px" | &nbsp; {{Vald flik|[[2.1 Inledning till integraler|Theory]]}} {{Ej vald flik|[[2.1 Övningar|Exercises]]}} | style="border-bottom:1px solid #797979" width="100%"| &nbsp; |} {{Info| '''Content:''' *Definition of an integral (overview). * Fundamental theorm of integral calculus. * Primitive function for <math>x^\alpha</math>, <math>1/x</math>, <math>e^x</math>, <math>\cos x</math> and <math>\sin x</math>. * Primitive function for sum and difference. }} {{Info| '''Learning outcomes:''' After this section, you will have learned to : * Interpret integrals as signed areas, that is, " the area above the <math>x</math>-axis" minus " area below the <math>x</math>-axis". * Understand other interpretations of the integral, for example. density / mass, speed / displacement, power / charge , etc. * 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. *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 <math>e^{x^2} </math>, <math>(\sin x)/x</math>, <math>\sin \sin x</math>, 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: {| width="100%" align="center" | width="2.5%" | | width="30%" |{{:2.1 - Figur - v-t-diagram med konstant fart 5}} | width="2.5%" | | width="30%" |{{:2.1 - Figur - v-t-diagram med konstant fart 4 och 6}} | width="2.5%" | | width="30%" |{{:2.1 - Figur - v-t-diagram med fart v(t) = t}} | width="2.5%" | |- || | valign="top" |<small> The object moves at a constant speed of 5.</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>The object is sliding down a sloping plane and has a linearly increasing speed. </small> || |} The distance travelled is in each case is {{Fristående formel||<math>s(6) = 5\cdot 6 = 30\,\mbox{m},\quad s(6) = 4\cdot 3 + 6\cdot 3 = 30\,\mbox{m},\quad s(6) = \frac{6\cdot 6}{2} = 18\,\mbox{m}\,\mbox{.}</math>}} 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. <div class="exempel"> ''' Example 1''' {| width="100%" align="center" | width="2.5%" | | width="30%" |{{:2.1 - Figur - Effekt-tid-diagram}} | width="2.5%" | | width="30%" |{{:2.1 - Figur - Kraft-sträcka-diagram}} | width="2.5%" | | width="30%" |{{:2.1 - Figur - Ström-tid-diagram}} | width="2.5%" | |- || | 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>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> A capacitor that is charged by a current ''i'' will receive a charge which is proportional to the area under the above graph. </small> || |} </div> == 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'' <math>\,\smallint\,</math> : <div class="tips"> 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>}} 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 class="exempel"> ''' Example 2''' {| width="100%" | 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 = \int_{a}^{\,c} f(x)\, dx\,\mbox{.}</math>}} | width="5%" | ||{{:2.1 - Figur - Area under grafen y = f(x) från a till b och c}} |} </div> <div class="exempel"> ''' Example 3''' {| width="100%" | 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>}} ''Note .'' We assume that speed and distance are measured using the same units of length. | width="5%" | ||{{:2.1 - Figur - Area s(10) i ett v-t-diagram}} |} </div> <div class="exempel"> ''' Example 4''' 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>}} specifies the amount in litres which flows into the tank during the tenth second. </div> <div class="exempel"> ''' Example 5''' Calculate the integrals {| width="100%" | width="95%" | <ol type="a"> <li> <math>\int_{0}^{4} 3 \, dx</math><br> <br> The integral can be interpreted as area below the curve (the line) <math>y=3</math> going from <math>x = 0</math> to <math>x = 4</math>, 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> </ol> | width="5%" | ||{{:2.1 - Figur - Area under grafen y = 3 från x = 0 till x = 4}} |} {| width="100%" | width="95%" | <ol type="a" start=2> <li><math>\int_{2}^{5} \Bigl(\frac{x}{2} -1 \Bigr) \, dx</math> <br><br> The integral can be interpreted as the area below the line <math>y=x/2-1</math> going from <math>x = 2</math> to <math>x = 5</math>, e. a triangle with a base 3 and a height <br> <center><math>\int_{2}^{5} \Bigl(\frac{x}{2} -1 \Bigr) \, dx = \frac{3 \cdot 1{,}5}{2} = 2{,}25\,\mbox{.}</math></center></li> </ol> | width="5%" | ||{{:2.1 - Figur - Area under grafen y = x/2 - 1 från x = 2 till x = 5}} |} {| width="100%" | width="95%" | <ol type="a" start=3> <li><math>\int_{0}^{a} kx \, dx\,\mbox{}\quad</math> där <math>k>0\,</math>.<br><br> The integral can be interpreted as the area below the line <math>y=kx</math> going from <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> <center><math>\int_{0}^{\,a} kx\,dx = \frac{a \cdot ka}{2} = \frac{ka^2}{2}\,\mbox{.}</math></center></li> </ol> | width="5%" | ||{{:2.1 - Figur - Area under grafen y = kx från x = 0 till x = a}} |} </div> == The primitive function == 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"> '''Exempel 6''' <ol type="a"> <li><math>F(x) = x^3 + \cos x - 5</math> is a primitive function of <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 = f(x)\,\mbox{.}</math>}}</li> <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 {{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> <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> </ol> </div> == 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 <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> {{Fristående formel||<math>A(x) = \int_{a}^{\,x} f(t) \, dt\,\mbox{.}</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> 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>}} 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>}} 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>}} Thus the function <math>A(x)</math> is a primitive function of <math>f(x)</math>. == Evaluating integrals == 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>}} 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>}} which gives that <math>C=-F(a)</math>. If we summarise, we have that {{Fristående formel||<math>\int_{a}^{\,b} f(t) \, dt = F(b) - F(a)\,\mbox{.}</math>}} 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 = F(b) - F(a)\,\mbox{.}</math>}} 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 = \Bigl[\,F(x)\,\Bigr]_{a}^{b} = F(b) - F(a)\,\mbox{.}</math>}} <div class="exempel"> ''' Example 7''' 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="95%" | {{Fristående formel||<math>\int_{0}^{2} (2x-x^2) \, dx\,\mbox{.}</math>}} 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>}} The area is<math>\frac{4}{3}</math>&nbsp;u.a. | width="5%" | ||{{: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. </div> == 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 <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>}} <div class="exempel"> ''' Example 8''' <ol type="a"> <li><math>\int (x^4 - 2x^3 + 4x - 7)\,dx = \frac{x^5}{5} - \frac{2x^4}{4} + \frac{4x^2}{2} - 7x + C</math><br/> <math>\phantom{\int (x^4 - 2x^3 + 4x - 7)\,dx}{} = \frac{x^5}{5} - \frac{x^4}{2} + 2x^2 - 7x + C</math></li> <li><math>\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</math><br> <math>\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(}</math></li> <li><math>\int \frac{2}{3x} \,dx = \int \frac{2}{3} \cdot \frac{1}{x} \, dx = \tfrac{2}{3} \ln |x| + C</math></li> <li><math>\int ( e^x - \cos x - \sin x ) \, dx = e^x - \sin x + \cos x +C</math></li> </ol> </div> == 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. <div class="exempel"> ''' Example 9''' <ol type="a"> <li><math>\int e^{3x} \, dx = \frac{e^{3x}}{3} + C</math></li> <li><math>\int \sin 5x \, dx = - \frac{ \cos 5x}{5} + C</math></li> <li><math>\int (2x +1)^4 \, dx = \frac{(2x+1)^5}{5 \cdot 2} + C</math></li> </ol> </div> <div class="exempel"> ''' Example 10''' <ol type="a"> <li><math>\int \sin kx \, dx = - \frac{\cos kx}{k} + C</math></li> <li><math>\int \cos kx \, dx = \frac{\sin kx }{k} + C</math></li> <li><math>\int e^{kx} \, dx = \displaystyle \frac{e^{kx}}{k} + C</math></li> </ol> </div> 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: # <math>\int_{b}^{\,a} f(x) \, dx = - \int_{a}^{\,b} f(x) \, dx\,\mbox{,}\vphantom{\Biggl(}</math> # <math>\int_{a}^{\,b} f(x) \, dx + \int_{a}^{\,b} g(x) \, dx = \int_{a}^{\,b} (f(x) + g(x)) \, dx\,\mbox{,}\vphantom{\Biggl(}</math> # <math>\int_{a}^{\,b} k \cdot f(x)\, dx = k \int_{a}^{\,b} f(x)\, dx\,\mbox{,}\vphantom{\Biggl(}</math> # <math>\int_{a}^{\,b} f(x) \, dx + \int_{b}^{\,c} f(x)\, dx = \int_{a}^{\,c} f(x)\, dx\,\mbox{.}</math> 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" ||<math>\begin{align*}A_1 &= \int_{a}^{\,b} f(x)\, dx,\\[6pt] A_2 &= -\int_{b}^{\,c} f(x)\, dx\,\mbox{.} \end{align*}</math> | width="10%" | ||{{:2.1 - Figur - Areor A₁ och A₂ mellan y = f(x) och x-axeln}} |} The total area is <math>\ A_1 + A_2 = \int_{a}^{\,b} f(x)\, dx - \int_{b}^{\,c} f(x)\, dx\,</math>. ''Note .'' The value of an '''integral''' can be negative, while an '''area''' always has a positive value. <div class="exempel"> ''' Example 11''' <ol type="a"> <li><math>\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</math><br> <math>\qquad{}= \Bigl[\,\tfrac{1}{4}x^4 - x^3 + x^2 + 3x\,\Bigr]_{1}^{2} \vphantom{\Biggr)^2}</math><br> <math>\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}</math><br> <math>\qquad{}=6-3-\tfrac{1}{4} = \tfrac{11}{4}</math><br/> {| | align="center" |{{:2.1 - Figur - Area för y = x³ - 3x² + 2x + 1, y = 2 och y = x³ - 3x² + 2x + 3}} |- ||<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> </ol> <ol type="a" start=2> <li><math>\int_{1}^{3} (x^2/2 - 2x) \, dx + \int_{1}^{3} (2x - x^2/2 + 3/2) \, dx = \int_{1}^{3} 3/2 \, dx</math><br/> <math>\qquad{} = \Bigl[\,\tfrac{3}{2}x\,\Bigr]_{1}^{3} = \tfrac{3}{2}\cdot 3 - \tfrac{3}{2}\cdot 1 = 3</math><br/> {| | align="center" |{{:2.1 - Figur - Area för y = x²/2 - 2x, y = 2x - x²/2 + 3/2 och y = 3/2}} |- ||<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> |} </li> </ol> <ol type="a" start=3> <li><math>\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(}</math><br> <math>\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(}</math><br> <math>\qquad{}= \frac {2}{3}\Bigl((4- \ln 2) - (1 - \ln 1)\Bigr) = \tfrac{2}{3}(3 - \ln 2) = 2 - \tfrac{2}{3}\ln 2 </math></li> </ol> <ol type="a" start=4> <li><math>\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</math><br/> {| | align="center" |{{:2.1 - Figur - Area för y = x² - 1}} |- ||<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> </ol> </div> == Area between curves == 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 - \int_{a}^{b} g(x) \, dx\,\mbox{,}</math>}} which can be simplified to {{Fristående formel||<math>\int_{a}^{b} (f(x) - g(x)) \, dx\,\mbox{.}</math>}} <center>{{:2.1 - Figur - Area mellan y = f(x) och y = g(x)}}</center> {| width="90%" align="center" ||<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> |} 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> {| width="90%" align="center" ||<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"> ''' Example 12''' 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> Since <math>e^x + 1 > 1 - x^2/2</math> in all the interval the area in question is given by {| width="100%" | width="95%" | {{Fristående formel||<math>\begin{align*} &\int_{-1}^{1} (e^x + 1) \, dx - \int_{-1}^{1} \Bigl( 1- \frac{x^2}{2}\Bigr) \, dx \vphantom{\Biggl(}\\ &\qquad{}= \int_{-1}^{1} \Bigl( e^x + \frac{x^2}{2} \Bigr) \, dx \vphantom{\Biggl(}\\ &\qquad{}= \Bigl[\,e^x + \frac{x^3}{6}\,\Bigr]_{-1}^{1} \vphantom{\Biggl(}\\ &\qquad{}= \Bigl( e^1 + \frac{1^3}{6} \Bigr) - \Bigl( e^{-1} + \frac{(-1)^3}{6} \Bigr)\vphantom{\Biggl(}\\ &\qquad{}= e - \frac{1}{e} + \frac{1}{3} \ \text{a.e.}\end{align*}</math>}} | width="5%" | ||{{:2.1 - Figur - Area mellan y = e^x - 1 och y = 1 - x²/2}} |} </div> <div class="exempel"> ''' Example 13''' 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> 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{or}\quad x=1\,\mbox{.}\end{align*}</math>}} {| width="100%" | 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}\\ &{}= \Bigl[\,\frac{3x^{4/3}}{4} - \frac{x^3}{3}\, \Bigr]_{0}^{1}\\[4pt] &{}= \tfrac{3}{4} - \tfrac{1}{3} - (0-0)\\[4pt] &{}= \tfrac{5}{12}\ \text{a.e.}\end{align*}</math>}} | width="5%" | ||{{:2.1 - Figur - Area mellan y = ∛x och y = x²}} |} </div> <div class="exempel"> ''' Example 14''' 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> {| width="100%" | width="95%" | 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 \quad\text{och}\quad A_2 = \int_{b}^{\,c} (2- x) \, dx\,\mbox{.}</math>}} We first determine the points of intersection <math>x=a</math>, <math>x=b</math> and <math>x=c</math>: | width="5%" | ||{{:2.1 - Figur - Area som begränsas av y = 1/x², y = x och y = 2}} |} * The point of intersection <math>x=a</math> is obtained from the equation {{Fristående formel||<math>\frac{1}{x^2} = 2 \quad \Leftrightarrow \quad x^2 = \frac{1}{2} \quad \Leftrightarrow \quad x = \pm \frac{1}{\sqrt{2}}\,\mbox{.}</math>}} :(The negative root, however, is not relevant.) * The point of intersection <math>x=b</math> is obtained from the equation {{Fristående formel||<math>\frac{1}{x^2} = x \quad \Leftrightarrow \quad x^3 = 1 \quad \Leftrightarrow \quad x=1\,\mbox{.}</math>}} *The point of intersection <math>x=c</math> is obtained from the equation <math>x = 2</math>. 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{.} \end{align*}</math>}} 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>}} </div>

• 2.1 - Figure - An area shifted in the y-direction
• 2.1 - Figure - Current-time-diagram
• 2.1 - Figure - Force-distance-diagram
• 2.1 - Figure - Power-time-diagram
• 2.1 - Figure - The area between y = e^x - 1 and y = 1 - x²/2
• 2.1 - Figure - The area between y = f(x) and y = g(x)
• 2.1 - Figure - The area between y = ∛x och y = x²
• 2.1 - Figure - The area bounded by y = 1/x², y = x and y = 2
• 2.1 - Figure - The area of y = x²/2 - 2x, y = 2x - x²/2 + 3/2 and y = 3/2
• 2.1 - Figure - The area of y = x² - 1
• 2.1 - Figure - The area of y = x³ - 3x² + 2x + 1, y = 2 and y = x³ - 3x² + 2x + 3
• 2.1 - Figure - The area s(10) in a v-t-diagram
• 2.1 - Figure - The area under the graph of y = 2x - x² from x = 0 to x = 2
• 2.1 - Figure - The area under the graph of y = 3 from x = 0 to x = 4
• 2.1 - Figure - The area under the graph of y = f(x) from a to b and c
• 2.1 - Figure - The area under the graph of y = f(x) from t = a to t = x
• 2.1 - Figure - The area under the graph of y = f(x) from t = a to t = x + h
• 2.1 - Figure - The area under the graph of y = kx from x = 0 to x = a
• 2.1 - Figure - The area under the graph of y = x/2 - 1 from x = 2 to x = 5
• 2.1 - Figure - The areas A₁ and A₂ between y = f(x) and the x-axis
• 2.1 - Figure - v-t-diagram with constant speed 4 and 6
• 2.1 - Figure - v-t-diagram with constant speed 5
• 2.1 - Figure - v-t-diagram with speed v(t) = t
• Template:Displayed math
• Template:Info
• Template:Not selected tab
• Template:Selected tab

Return to Main Page.