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I have this integral: $\displaystyle \int^{\infty}_0 kx e^{-kx} dx$.

I tried integrating it by parts:

$\dfrac{1}{k}\displaystyle \int^{\infty}_0 kx e^{-kx} dx = ... $. But I'm stuck

now. Can you help me please?

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    $\begingroup$ Where are you stuck? Can you say what you have tried? $\endgroup$ – Susan_Math123 Sep 4 '17 at 5:39
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hint: $\displaystyle \int kxe^{-kx}dx = -\displaystyle \int xd(e^{-kx})$

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  • $\begingroup$ @Yang-Hui I would like to clarify this answer for you a bit more. DeepSea is recommending you make the substitution $u=e^{-kx}$, making $$du=d(e^{-kx})=-ke^{-kx}\,dx$$ $\endgroup$ – gen-z ready to perish Sep 4 '17 at 6:24
  • $\begingroup$ Am I right @DeepSea ? $\endgroup$ – gen-z ready to perish Sep 4 '17 at 6:26
  • $\begingroup$ @ChaseRyanTaylor: You got it Chase ! $\endgroup$ – DeepSea Sep 4 '17 at 6:26
  • $\begingroup$ $$\Huge\ddot\smile$$ $\endgroup$ – gen-z ready to perish Sep 4 '17 at 6:29
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Just take $kx = y$ so you will have $$\int_{0}^\infty\frac{y}{e^y}dy$$

Take $u = y$ and $dv = \frac{1}{e^y}$, after integrate just remeber change $y = kx$

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Try explicitly writing out what $u$ and $dv$ are in your integration by parts. In this case, you should have $u=kx$, and $dv=e^{-kx}\,dx$. Thus, $du=k\,dx$, and $v=-\frac{1}{k}e^{-kx}$. Now apply the integration-by-parts formula:

$$\int_{0}^{\infty}{u\,dv}={\left.uv\right|}_0^{\infty}-\int_{0}^{\infty}{v\,du}$$

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