How do you integrate $\int \frac{y}{e^{2 y}}$ $int y/(e^(2y))$ by integration by parts method?

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How do you integrate $\int \frac{y}{e^{2 y}}$ $int y/(e^(2y))$ by integration by parts method?

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Answer 1 · 2016-12-09T16:15:32

$I = c - \frac{y e^{- 2 y}}{2} - \frac{e^{- 2 y}}{4}$ $I = c -(ye^(-2y))/2 - e^(-2y)/4$

Explanation:

We have

$I = \int \frac{y}{e^{2 y}} d y$ $I = int y/e^(2y)dy$

With some simple rewriting, based on the properties of exponentials $\frac{1}{a^{x}} = a^{- x}$ $1/a^(x) = a^(-x)$

$I = \int y e^{- 2 y} d y$ $I = int ye^(-2y)dy$

If we say

$u = y$ $u = y$ so $d u = d y$ $du = dy$
$d v = e^{- 2 y} d y$ $dv = e^(-2y)dy$ so $v = - \frac{e^{- 2 y}}{2}$ $v = -e^(-2y)/2$

By the integration by parts formula

$\int u d v = u v - \int v d u$ $int udv = uv - int vdu$

$I = - \frac{y e^{- 2 y}}{2} - \int - \frac{e^{- 2 y}}{2} d y$ $I = -(ye^(-2y))/2 - int-e^(-2y)/2dy$

Putting the constants outside of the integral

$I = - \frac{y e^{- 2 y}}{2} + \frac{1}{2} \int e^{- 2 y} d y$ $I = -(ye^(-2y))/2 +1/2int e^(-2y)dy$
$I = c - \frac{y e^{- 2 y}}{2} - \frac{e^{- 2 y}}{4}$ $I = c -(ye^(-2y))/2 - e^(-2y)/4$

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How do you integrate $\int \frac{y}{e^{2 y}}$ $int y/(e^(2y))$ by integration by parts method?

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Explanation:

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