Evaluate $\int {x}^{2}{e}^{-{x}^{3}}dx.$

Aidyn Crosby
2022-09-30
Answered

Evaluate $\int {x}^{2}{e}^{-{x}^{3}}dx.$

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ralharn

Answered 2022-10-01
Author has **15** answers

Given $\int {x}^{2}{e}^{-{x}^{3}}dx.$

Applying u Substitution

$u=-{x}^{3}\phantom{\rule{0ex}{0ex}}\frac{du}{dx}=-3{x}^{2}\phantom{\rule{0ex}{0ex}}dx=\frac{-du}{3{x}^{2}}$

Now $\int {x}^{2}{e}^{-{x}^{3}}dx=\phantom{\rule{0ex}{0ex}}=\frac{-1}{3}\int {e}^{u}du\phantom{\rule{0ex}{0ex}}=\frac{-{e}^{u}}{3}$

Now, pot $u=-{x}^{3}$, and add constant

$=\frac{-1}{3}{e}^{-{x}^{3}}+C$

Applying u Substitution

$u=-{x}^{3}\phantom{\rule{0ex}{0ex}}\frac{du}{dx}=-3{x}^{2}\phantom{\rule{0ex}{0ex}}dx=\frac{-du}{3{x}^{2}}$

Now $\int {x}^{2}{e}^{-{x}^{3}}dx=\phantom{\rule{0ex}{0ex}}=\frac{-1}{3}\int {e}^{u}du\phantom{\rule{0ex}{0ex}}=\frac{-{e}^{u}}{3}$

Now, pot $u=-{x}^{3}$, and add constant

$=\frac{-1}{3}{e}^{-{x}^{3}}+C$

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