From Förberedande kurs i matematik 2

Revision as of 14:46, 16 September 2008 (edit) Tekbot (Talk | contribs) m (Lösning 2.3:1b moved to Solution 2.3:1b: Robot: moved page) ← Previous diff		Current revision (08:13, 29 October 2008) (edit) (undo) Tek (Talk | contribs) m
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-	{{NAVCONTENT_START}}	+	If we look at the formula for integration by parts,
-	<center> [[Image:2_3_1b.gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{Displayed math||<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!
		+
		+	{{Displayed math||<math>\begin{align}
		+	\int (x+1)\sin x\,dx
		+	&= (x+1)\cdot (-\cos x) - \int 1\cdot (-\cos x)\,dx\\[5pt]
		+	&= -(x+1)\cos x + \int \cos x\,dx\\[5pt]
		+	&= -(x+1)\cos x + \sin x + C\,\textrm{.}
		+	\end{align}</math>}}

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Current revision

If we look at the formula for integration by parts,

\displaystyle \int f(x)g(x)\,dx = F(x)g(x) - \int F(x)g'(x)\,dx\,,

we see that if we choose \displaystyle f(x)=\sin x and \displaystyle g(x)=x+1, then the factor \displaystyle 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 \displaystyle f(x) (which we can) and that we can then integrate it. Let's try!

\displaystyle \begin{align} \int (x+1)\sin x\,dx &= (x+1)\cdot (-\cos x) - \int 1\cdot (-\cos x)\,dx\\[5pt] &= -(x+1)\cos x + \int \cos x\,dx\\[5pt] &= -(x+1)\cos x + \sin x + C\,\textrm{.} \end{align}