How do you integrate $\int 2 x e^{x} d x$ $int2xe^x dx$ from 0 to 1?

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How do you integrate $\int 2 x e^{x} d x$ $int2xe^x dx$ from 0 to 1?

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Answer 1 · 2015-04-05T21:03:38

By parts:

$\int_{a}^{b} 2 x e^{x} d x = u v - \int_{a}^{b} v d u$ $int_a^b 2xe^xdx = uv - int_a^bvdu$

Let u = $2 x$ $2x$ and v = $e^{x}$ $e^x$ .
$d u$ $du$ = $2 d x$ $2dx$

$\int_{0}^{1} 2 x e^{x} d x = (2 x \cdot e^{x}) - \int_{0}^{1} e^{x} \cdot 2 d x$ $int_0^1 2xe^x dx = (2x*e^x) - int_0^1e^x*2 dx$
$= [(2 x \cdot e^{x}) - 2 \int_{0}^{1} e^{x} d x] e v a l (0 \to 1)$ $= [(2x*e^x) - 2int_0^1e^x dx] eval (0->1)$
$= [2 (1) e^{1} - 2 (0) e^{0}] - 2 (e^{1} - e^{0})$ $= [2(1)e^(1) - 2(0)e^(0)] - 2(e^1 - e^0)$
$= [2 e] - 2 e + 2$ $= [2e] - 2e + 2$
$= 2$ $= 2$

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