How do you integrate $\int x e^{- \frac{x}{2}}$ $int xe^(-x/2)$ by parts from $[0, 4]$ $[0,4]$ ?

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How do you integrate $\int x e^{- \frac{x}{2}}$ $int xe^(-x/2)$ by parts from $[0, 4]$ $[0,4]$ ?

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Answer 1 · 2018-03-22T01:38:13

$- 4 (3 e^{- 2} - 1)$ $-4(3e^-2-1)$

Explanation:

Let's select our values of $u, d v$ $u, dv$ and integrate and differentiate for $d u, v$ $du, v$ :

$u = x$ $u=x$
$d u = d x$ $du=dx$

$d v = e^{- \frac{x}{2}} d x$ $dv=e^(-x/2)dx$
$v = - 2 e^{- \frac{x}{2}}$ $v=-2e^(-x/2)$

Note that our bounds of integration will not change -- we're not performing any substitutions.

So, for definite integrals, we integrate by parts in the following way:

$\int_{a}^{b} u d v = u v ∣_{a}^{b} - \int_{a}^{b} v d u$ $int_a^budv=uv|_a^b-int_a^bvdu$

Thus,

$\int_{0}^{4} x e^{- \frac{x}{2}} d x = - 2 x e^{- \frac{x}{2}} ∣ ∣ ∣_0^{4} + 2 \int_{0}^{4} \frac{e^{- x}}{2} d x = (- 8 e^{- 2}) - 4 e^{- (\frac{x}{2})} ∣ ∣ ∣ 0^{4} = - 8 e^{- 2} - 4 e^{- 2} + 4 = - 12 e^{- 2} + 4 = - 4 (3 e^{- 2} - 1)$ $int_0^4xe^(-x/2)dx=-2xe^(-x/2)|_0^4 +2int_0^4e^-x/2dx=(-8e^-2)-4e^-(x/2)|0^4=-8e^-2-4e^-2+4=-12e^-2+4=-4(3e^-2-1)$

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How do you integrate $\int x e^{- \frac{x}{2}}$ $int xe^(-x/2)$ by parts from $[0, 4]$ $[0,4]$ ?

1 Answer

Explanation:

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How do you integrate $\int x e^{- \frac{x}{2}}$ $int xe^(-x/2)$ by parts from $[0, 4]$ $[0,4]$ ?