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Q 35.

Expert-verifiedFound in: Page 429

Book edition
1st

Author(s)
Peter Kohn, Laura Taalman

Pages
1155 pages

ISBN
9781429241861

Solve the integral: $\int \frac{x}{{e}^{x}}dx$

The required answer is $-x{e}^{x}-{e}^{x}+c$.

We have given integral is $\int \frac{x}{{e}^{x}}dx$.

We have, $u=x,dv=\frac{dx}{{e}^{x}}$

$u=x\phantom{\rule{0ex}{0ex}}du=dx$

and

$dv=\frac{dx}{{e}^{x}}\phantom{\rule{0ex}{0ex}}v=\int \frac{dx}{{e}^{x}}\phantom{\rule{0ex}{0ex}}v=-\frac{1}{{e}^{x}}$

The formula of integration by parts is $\int udv=uv-\int vdu$

$\int \frac{x}{{e}^{x}}dx\phantom{\rule{0ex}{0ex}}=\left(x\left(-\frac{1}{{e}^{x}}\right)-\int \left(-\frac{1}{{e}^{x}}\right)dx\right)\phantom{\rule{0ex}{0ex}}=\left(-\frac{x}{{e}^{x}}\right)+\int {e}^{-x}dx$

We have, $u=-x\phantom{\rule{0ex}{0ex}}du=-dx$

So,

$\left(-\frac{x}{{e}^{x}}\right)+\int {e}^{-x}du\phantom{\rule{0ex}{0ex}}=\left(-\frac{x}{{e}^{x}}\right)-{e}^{-x}+c\phantom{\rule{0ex}{0ex}}=-x{e}^{x}-{e}^{x}+c$

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