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Question #5a569

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Dec 25, 2016

$\int x^{3} e^{x} d x = e^{x} (x^{3} - 3 x^{2} + 6 x - 6)$ $int x^3e^xdx = e^x(x^3 -3x^2+6x-6)$

Explanation:

You can calculate the integral by parts iteratively:

$\int x^{3} e^{x} d x = \int x^{3} d (e^{x}) = x^{3} e^{x} - \int 3 x^{2} e^{x} d x$ $int x^3e^xdx= int x^3 d(e^x) = x^3e^x-int 3x^2e^xdx$

$\int x^{2} e^{x} d x = x^{2} e^{x} - 2 \int x e^{x} d x$ $int x^2e^xdx = x^2e^x-2 int xe^xdx$

$\int x e^{x} d x = x e^{x} - \int e^{x} d x = e^{x} (x - 1)$ $int xe^xdx= xe^x - int e^xdx = e^x(x-1)$

Putting it all together:

$\int x^{3} e^{x} d x = e^{x} (x^{3} - 3 x^{2} + 6 x - 6)$ $int x^3e^xdx = e^x(x^3 -3x^2+6x-6)$

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