# Problem integrating by parts $e^{-2x}$

I know this question is very basic but I can't really make it..

$\int{e^{-2x}dx}=?$

\begin{align} u = -2x, &du= -2 dx\\ v = e, &dv= e dx\end{align}

then applying the rules $uv-\int vdu$ $$-2xe - \int -2e dx = -2ex + 2 \int edx = -2ex + 2e$$ but the derivate of $-2ex + 2e$ is equal to $-2e$

What is wrong? thanks

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This isn't an integration by parts problem; it's substitution. Set $u=-2x$ and $du=-2\,dx$. You wind up integrating $(-1/2)e^u\,du$. –  David Mitra Nov 18 '11 at 11:06
You're confusing $v^u$ with $v \cdot u$. –  Hans Lundmark Nov 18 '11 at 11:12

What is wrong? Where to begin....

If $v=e$, then $dv\ne e\,dx$; $e$ is a constant, so $dv=0\,dx$.

The left side of the equation for integration by parts is $\int u\,dv$. If $u=-2x$, and $dv = e\,dx$, then $u\,dv=-2ex\,dx$, which is not $e^{-2x}$, which is what you want.

And the integral of $e$ isn't $e$, it's $ex+C$.

Finally, as David Mitra points out in the comments, the problem is better done by other means.

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I've been thinking that as derivate_of(e^x)=e^x, then derivate_of(e)=e, but this is wrong. –  Totty Nov 18 '11 at 12:24

It's easier to first differentiate $e^{-2x}$: $$\frac{d}{dx}e^{-2x}=-2e^{-2x}.$$

This means $$\int -2e^{-2x}\,dx=e^{-2x}+C,$$ so dividing by $(-2)$ we get: $$\int e^{-2x}\,dx=-\frac12e^{-2x}+C'\quad (C': \text{constant}).$$

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You aren't applying integration by parts correctly. Integration by parts is used to find the antiderivative of a product of functions. The product $u\, dv$ has to be the integrand.

Your function, $f(x)=e^{-2x}$ is a composition of functions. You can use the $u$-substitution method to find the antiderivative.

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I can only do by parts... –  Totty Nov 18 '11 at 12:15