Lösung 2.1:2a - Versionsgeschichte

Sekretariat1 um 07:55, 21. Aug. 2009

Sekretariat1 — Fri, 21 Aug 2009 07:55:33 GMT

← Nächstältere Version		Version vom 07:55, 21. Aug. 2009
Zeile 25:		Zeile 25:
	F'(x) &= \tfrac{1}{3}\bigl(x^3\bigr)' + \tfrac{3}{4}\bigl(x^4\bigr)'\\[5pt]		F'(x) &= \tfrac{1}{3}\bigl(x^3\bigr)' + \tfrac{3}{4}\bigl(x^4\bigr)'\\[5pt]
	&= \tfrac{1}{3}\cdot 3x^2 + \tfrac{3}{4}\cdot 4x^3\\[5pt]		&= \tfrac{1}{3}\cdot 3x^2 + \tfrac{3}{4}\cdot 4x^3\\[5pt]
-	&= x^2+3x^3\,.	+	&= x^2+3x^3\,\textrm{.}
	\end{align}</math>}}		\end{align}</math>}}

Sekretariat1 um 07:51, 21. Aug. 2009

Sekretariat1 — Fri, 21 Aug 2009 07:51:52 GMT

← Nächstältere Version		Version vom 07:51, 21. Aug. 2009
Zeile 1:		Zeile 1:
-	Die Schwierigkeit in der Integralrechnung, liegt darin eine Stammfunktion zu finden. Danach müssen wir nur die Stammfunktion in den beiden Integrationzgrenzen evaluieren	+	Die Schwierigkeit in der Integralrechnung liegt darin, eine Stammfunktion zu finden. Danach müssen wir nur die Stammfunktion in den beiden Integrationzgrenzen evaluieren.

-	~~Nachdem~~ unser Integrand ~~auf~~ der Form <math>x^n</math> ist, können wir die Regel	+	Da unser Integrand in der Form <math>x^n</math> ist, können wir die Regel

	{{Abgesetzte Formel\|\|<math>\int x^n\,dx = \frac{x^{n+1}}{n+1}+C</math>}}		{{Abgesetzte Formel\|\|<math>\int x^n\,dx = \frac{x^{n+1}}{n+1}+C</math>}}

-	für jeden Term benutzen,	+	für jeden Term benutzen.

	{{Abgesetzte Formel\|\|<math>F(x) = \frac{x^{2+1}}{2+1} + 3\cdot \frac{x^{3+1}}{3+1}</math>}}		{{Abgesetzte Formel\|\|<math>F(x) = \frac{x^{2+1}}{2+1} + 3\cdot \frac{x^{3+1}}{3+1}</math>}}
Zeile 20:		Zeile 20:


-	Hinweis: Wir können testen ob <math>F(x) = \tfrac{1}{3}x^3 + \tfrac{3}{4}x^4</math> eine Stammfunktion von ~~den~~ Integrand ist, indem wir <math>F(x)</math> ableiten	+	Hinweis: Wir können testen ob <math>F(x) = \tfrac{1}{3}x^3 + \tfrac{3}{4}x^4</math> eine Stammfunktion von dem Integrand ist, indem wir <math>F(x)</math> ableiten

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
	F'(x) &= \tfrac{1}{3}\bigl(x^3\bigr)' + \tfrac{3}{4}\bigl(x^4\bigr)'\\[5pt]		F'(x) &= \tfrac{1}{3}\bigl(x^3\bigr)' + \tfrac{3}{4}\bigl(x^4\bigr)'\\[5pt]
	&= \tfrac{1}{3}\cdot 3x^2 + \tfrac{3}{4}\cdot 4x^3\\[5pt]		&= \tfrac{1}{3}\cdot 3x^2 + \tfrac{3}{4}\cdot 4x^3\\[5pt]
-	&= x^2+3x^3	+	&= x^2+3x^3\,.
	\end{align}</math>}}		\end{align}</math>}}

Martin um 16:40, 28. Apr. 2009

Martin — Tue, 28 Apr 2009 16:40:04 GMT

← Nächstältere Version		Version vom 16:40, 28. Apr. 2009
Zeile 1:		Zeile 1:
-	The foremost difficulty with calculating an integral is finding a primitive function of the integrand. Once we have done that, the integral is calculated as the difference between the primitive function's values in ~~the upper and lower limits of integration~~.	+	Die Schwierigkeit in der Integralrechnung, liegt darin eine Stammfunktion zu finden. Danach müssen wir nur die Stammfunktion in den beiden Integrationzgrenzen evaluieren

-	~~The integrand in our case consists of two terms in the form~~ <math>x^n</math>, ~~and so we can use the rule~~	+	Nachdem unser Integrand auf der Form <math>x^n</math> ist, können wir die Regel

	{{Abgesetzte Formel\|\|<math>\int x^n\,dx = \frac{x^{n+1}}{n+1}+C</math>}}		{{Abgesetzte Formel\|\|<math>\int x^n\,dx = \frac{x^{n+1}}{n+1}+C</math>}}

-	~~on the terms individually to obtain that~~	+	für jeden Term benutzen,

	{{Abgesetzte Formel\|\|<math>F(x) = \frac{x^{2+1}}{2+1} + 3\cdot \frac{x^{3+1}}{3+1}</math>}}		{{Abgesetzte Formel\|\|<math>F(x) = \frac{x^{2+1}}{2+1} + 3\cdot \frac{x^{3+1}}{3+1}</math>}}

-	~~is a primitive function of the integrand.~~	+	Der Wert des Integrals ist daher
-		+
-	~~The integrand's value is thus~~	+

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 22:		Zeile 20:


-	~~Note~~: ~~One way to check that~~ <math>F(x) = \tfrac{1}{3}x^3 + \tfrac{3}{4}x^4</math> ~~is a primitive function of the integral is to differentiate~~ <math>F(x)</math> ~~and to see that we obtain~~	+	Hinweis: Wir können testen ob <math>F(x) = \tfrac{1}{3}x^3 + \tfrac{3}{4}x^4</math> eine Stammfunktion von den Integrand ist, indem wir <math>F(x)</math> ableiten

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 29:		Zeile 27:
	&= x^2+3x^3		&= x^2+3x^3
	\end{align}</math>}}		\end{align}</math>}}
-
-	as the integrand.

Tekbot: Solution 2.1:2a moved to Lösung 2.1:2a: Robot: moved page

Tekbot — Wed, 11 Mar 2009 10:15:15 GMT

Solution 2.1:2a moved to Lösung 2.1:2a: Robot: moved page

← Nächstältere Version

Version vom 10:15, 11. Mär. 2009

Tekbot: Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel)

Tekbot — Tue, 10 Mar 2009 12:58:04 GMT

Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel)

← Nächstältere Version		Version vom 12:58, 10. Mär. 2009
Zeile 3:		Zeile 3:
	The integrand in our case consists of two terms in the form <math>x^n</math>, and so we can use the rule		The integrand in our case consists of two terms in the form <math>x^n</math>, and so we can use the rule

-	{{~~Displayed math~~\|\|<math>\int x^n\,dx = \frac{x^{n+1}}{n+1}+C</math>}}	+	{{Abgesetzte Formel\|\|<math>\int x^n\,dx = \frac{x^{n+1}}{n+1}+C</math>}}

	on the terms individually to obtain that		on the terms individually to obtain that

-	{{~~Displayed math~~\|\|<math>F(x) = \frac{x^{2+1}}{2+1} + 3\cdot \frac{x^{3+1}}{3+1}</math>}}	+	{{Abgesetzte Formel\|\|<math>F(x) = \frac{x^{2+1}}{2+1} + 3\cdot \frac{x^{3+1}}{3+1}</math>}}

	is a primitive function of the integrand.		is a primitive function of the integrand.
Zeile 13:		Zeile 13:
	The integrand's value is thus		The integrand's value is thus

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	\int\limits_{0}^{2} \bigl( x^2+3x^3\bigr)\,dx		\int\limits_{0}^{2} \bigl( x^2+3x^3\bigr)\,dx
	&= \Bigl[\ \frac{x^3}{3} + 3\cdot\frac{x^4}{4}\Bigr]_0^2\\		&= \Bigl[\ \frac{x^3}{3} + 3\cdot\frac{x^4}{4}\Bigr]_0^2\\
Zeile 24:		Zeile 24:
	Note: One way to check that <math>F(x) = \tfrac{1}{3}x^3 + \tfrac{3}{4}x^4</math> is a primitive function of the integral is to differentiate <math>F(x)</math> and to see that we obtain		Note: One way to check that <math>F(x) = \tfrac{1}{3}x^3 + \tfrac{3}{4}x^4</math> is a primitive function of the integral is to differentiate <math>F(x)</math> and to see that we obtain

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	F'(x) &= \tfrac{1}{3}\bigl(x^3\bigr)' + \tfrac{3}{4}\bigl(x^4\bigr)'\\[5pt]		F'(x) &= \tfrac{1}{3}\bigl(x^3\bigr)' + \tfrac{3}{4}\bigl(x^4\bigr)'\\[5pt]
	&= \tfrac{1}{3}\cdot 3x^2 + \tfrac{3}{4}\cdot 4x^3\\[5pt]		&= \tfrac{1}{3}\cdot 3x^2 + \tfrac{3}{4}\cdot 4x^3\\[5pt]

Tek um 12:34, 21. Okt. 2008

Tek — Tue, 21 Oct 2008 12:34:52 GMT

← Nächstältere Version		Version vom 12:34, 21. Okt. 2008
Zeile 1:		Zeile 1:
	The foremost difficulty with calculating an integral is finding a primitive function of the integrand. Once we have done that, the integral is calculated as the difference between the primitive function's values in the upper and lower limits of integration.		The foremost difficulty with calculating an integral is finding a primitive function of the integrand. Once we have done that, the integral is calculated as the difference between the primitive function's values in the upper and lower limits of integration.

-	The integrand in our case consists of two terms in the form	+	The integrand in our case consists of two terms in the form <math>x^n</math>, and so we can use the rule
-	<math>x^{n}</math>, and so we can use the rule	+
-		+
-		+
-	~~<math>\int{x^{n}\,dx=\frac{x^{n+1}}{n+1}}+C</math>~~	+

		+	{{Displayed math\|\|<math>\int x^n\,dx = \frac{x^{n+1}}{n+1}+C</math>}}

	on the terms individually to obtain that		on the terms individually to obtain that

-		+	{{Displayed math\|\|<math>F(x) = \frac{x^{2+1}}{2+1} + 3\cdot \frac{x^{3+1}}{3+1}</math>}}
-	<math>F~~\left~~( x ~~\right~~)=\frac{x^{2+1}}{2+1}+3\~~centerdot~~ \frac{x^{3+1}}{3+1}</math>	+
-		+

	is a primitive function of the integrand.		is a primitive function of the integrand.
Zeile 18:		Zeile 13:
	The integrand's value is thus		The integrand's value is thus

		+	{{Displayed math\|\|<math>\begin{align}
		+	\int\limits_{0}^{2} \bigl( x^2+3x^3\bigr)\,dx
		+	&= \Bigl[\ \frac{x^3}{3} + 3\cdot\frac{x^4}{4}\Bigr]_0^2\\
		+	&= \frac{2^3}{3} + 3\cdot\frac{2^4}{4} - \Bigl(\frac{0^3}{3} + 3\cdot\frac{0^4}{4} \Bigr)\\[5pt]
		+	&= \frac{8}{3} + \frac{3\cdot 16}{4}\\[5pt]
		+	&= \frac{44}{3}\,\textrm{.}
		+	\end{align}</math>}}

-	<math>\begin{align}
-	& \int\limits_{0}^{2}{\left( x^{2}+3x^{3} \right)}\,dx=\left[ \frac{x^{3}}{3}+3\centerdot \frac{x^{4}}{4} \right]_{0}^{2} \\
-	& =\frac{2^{3}}{3}+3\centerdot \frac{2^{4}}{4}-\left( \frac{0^{3}}{3}+3\centerdot \frac{0^{4}}{4} \right) \\
-	& =\frac{8}{3}+\frac{3\centerdot 16}{4}=\frac{44}{3} \\
-	\end{align}</math>
-
-	NOTE: One way to check that
-	<math>F\left( x \right)=\frac{1}{3}x^{3}+\frac{3}{4}x^{4}</math>
-	is a primitive function of the integral is to differentiate
-	<math>F\left( x \right)</math>
-	and to see that we obtain

		+	Note: One way to check that <math>F(x) = \tfrac{1}{3}x^3 + \tfrac{3}{4}x^4</math> is a primitive function of the integral is to differentiate <math>F(x)</math> and to see that we obtain

-	<math>\begin{align}	+	{{Displayed math\|\|<math>\begin{align}
-	~~& {~~F}'~~\left~~( x ~~\right~~)=\~~frac~~{1}{3}\~~left~~( x^{3} \~~right~~)~~^{\prime }~~+\~~frac~~{3}{4}\~~left~~( x^{4} \~~right~~)~~^{\prime }~~ \\	+	F'(x) &= \tfrac{1}{3}\bigl(x^3\bigr)' + \tfrac{3}{4}\bigl(x^4\bigr)'\\[5pt]
-	& =\~~frac~~{1}{3}\~~centerdot~~ 3x^{2}+\~~frac~~{3}{4}\~~centerdot~~ 4x^{3}=x^{2}+3x^{3~~} \\~~	+	&= \tfrac{1}{3}\cdot 3x^2 + \tfrac{3}{4}\cdot 4x^3\\[5pt]
-	\end{align}</math>	+	&= x^2+3x^3
		+	\end{align}</math>}}

	as the integrand.		as the integrand.

Ian um 12:44, 17. Okt. 2008

Ian — Fri, 17 Oct 2008 12:44:44 GMT

← Nächstältere Version		Version vom 12:44, 17. Okt. 2008
Zeile 1:		Zeile 1:
-	{{~~NAVCONTENT_START~~}}	+	The foremost difficulty with calculating an integral is finding a primitive function of the integrand. Once we have done that, the integral is calculated as the difference between the primitive function's values in the upper and lower limits of integration.
-	<~~center~~> ~~[[Image:2_1_2a-~~1(2)~~.gif]~~] </~~center~~>	+
-	{{~~NAVCONTENT_STOP~~}}	+	The integrand in our case consists of two terms in the form
-	{{~~NAVCONTENT_START~~}}	+	<math>x^{n}</math>, and so we can use the rule
-	<~~center~~> ~~[[Image:2_1_2a-2~~(2)~~.gif]]~~ </~~center~~>	+
-	{{~~NAVCONTENT_STOP~~}}	+
		+	<math>\int{x^{n}\,dx=\frac{x^{n+1}}{n+1}}+C</math>
		+
		+
		+	on the terms individually to obtain that
		+
		+
		+	<math>F\left( x \right)=\frac{x^{2+1}}{2+1}+3\centerdot \frac{x^{3+1}}{3+1}</math>
		+
		+
		+	is a primitive function of the integrand.
		+
		+	The integrand's value is thus
		+
		+
		+	<math>\begin{align}
		+	& \int\limits_{0}^{2}{\left( x^{2}+3x^{3} \right)}\,dx=\left[ \frac{x^{3}}{3}+3\centerdot \frac{x^{4}}{4} \right]_{0}^{2} \\
		+	& =\frac{2^{3}}{3}+3\centerdot \frac{2^{4}}{4}-\left( \frac{0^{3}}{3}+3\centerdot \frac{0^{4}}{4} \right) \\
		+	& =\frac{8}{3}+\frac{3\centerdot 16}{4}=\frac{44}{3} \\
		+	\end{align}</math>
		+
		+	NOTE: One way to check that
		+	<math>F\left( x \right)=\frac{1}{3}x^{3}+\frac{3}{4}x^{4}</math>
		+	is a primitive function of the integral is to differentiate
		+	<math>F\left( x \right)</math>
		+	and to see that we obtain
		+
		+
		+	<math>\begin{align}
		+	& {F}'\left( x \right)=\frac{1}{3}\left( x^{3} \right)^{\prime }+\frac{3}{4}\left( x^{4} \right)^{\prime } \\
		+	& =\frac{1}{3}\centerdot 3x^{2}+\frac{3}{4}\centerdot 4x^{3}=x^{2}+3x^{3} \\
		+	\end{align}</math>
		+
		+	as the integrand.

Tekbot: Lösning 2.1:2a moved to Solution 2.1:2a: Robot: moved page

Tekbot — Tue, 16 Sep 2008 14:35:41 GMT

Lösning 2.1:2a moved to Solution 2.1:2a: Robot: moved page

← Nächstältere Version

Version vom 14:35, 16. Sep. 2008

Tekbot: Robot: Automated text replacement (-[[Bild: +[[Image:)

Tekbot — Thu, 21 Aug 2008 07:57:34 GMT

Robot: Automated text replacement (-[[Bild: +[[Image:)

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Oskar — Mon, 07 Apr 2008 08:07:44 GMT

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