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Lösung 2.3:1b

Aus Online Mathematik Brückenkurs 2

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If we look at the formula for integration by parts,
If we look at the formula for integration by parts,
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{{Displayed math||<math>\int f(x)g(x)\,dx = F(x)g(x) - \int F(x)g'(x)\,dx\,,</math>}}
+
{{Abgesetzte Formel||<math>\int f(x)g(x)\,dx = F(x)g(x) - \int F(x)g'(x)\,dx\,,</math>}}
we see that if we choose <math>f(x)=\sin x</math> and <math>g(x)=x+1</math>, then the factor <math>g(x)</math> will be differentiated to a constant on the right-hand side of the integral. Naturally, this presupposes that we can find a primitive function for <math>f(x)</math> (which we can) and that we can then integrate it. Let's try!
we see that if we choose <math>f(x)=\sin x</math> and <math>g(x)=x+1</math>, then the factor <math>g(x)</math> will be differentiated to a constant on the right-hand side of the integral. Naturally, this presupposes that we can find a primitive function for <math>f(x)</math> (which we can) and that we can then integrate it. Let's try!
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{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
\int (x+1)\sin x\,dx
\int (x+1)\sin x\,dx
&= (x+1)\cdot (-\cos x) - \int 1\cdot (-\cos x)\,dx\\[5pt]
&= (x+1)\cdot (-\cos x) - \int 1\cdot (-\cos x)\,dx\\[5pt]

Version vom 13:03, 10. Mär. 2009

If we look at the formula for integration by parts,

f(x)g(x)dx=F(x)g(x)−F(x)g(x)dx 

we see that if we choose f(x)=sinx and g(x)=x+1, then the factor g(x) will be differentiated to a constant on the right-hand side of the integral. Naturally, this presupposes that we can find a primitive function for f(x) (which we can) and that we can then integrate it. Let's try!

(x+1)sinxdx=(x+1)(−cosx)−1(−cosx)dx=−(x+1)cosx+cosxdx=−(x+1)cosx+sinx+C.
Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_2.3:1b“