Evaluate the following definite integrals: int_0^1(x e^(-x^2+2))dx

Question
Applications of integrals
asked 2021-02-19
Evaluate the following definite integrals:
\(\displaystyle{\int_{{0}}^{{1}}}{\left({x}{e}^{{-{x}^{{2}}+{2}}}\right)}{\left.{d}{x}\right.}\)

Answers (1)

2021-02-20
Step 1
To Evaluate the following definite integrals:
Step 2
Given That
\(\displaystyle{\int_{{0}}^{{1}}}{x}{e}^{{-{x}^{{2}}+{2}}}{\left.{d}{x}\right.}\)
Let \(\displaystyle{e}^{{-{x}^{{2}}+{2}}}={u}\)
\(\displaystyle{d}\frac{{u}}{{\left.{d}{x}\right.}}=-{2}{x}{\left({e}^{{-{x}^{{2}}+{2}}}\right)}\)
\(\displaystyle{d}\frac{{u}}{{-{{2}}}}={x}{e}^{{-{x}^{{2}}+{2}}}{\left.{d}{x}\right.}\)
Hence \(\displaystyle{\int_{{0}}^{{1}}}{x}{e}^{{-{x}^{{2}}+{2}}}{\left.{d}{x}\right.}={\int_{{0}}^{{1}}}{d}\frac{{u}}{{-{{2}}}}\)
\(\displaystyle=-\frac{{1}}{{2}}{{\left[{4}\right]}_{{0}}^{{1}}}\)
Since we have \(\displaystyle{u}={e}^{{-{x}^{{2}}+{2}}}\) Then we have
\(\displaystyle=-\frac{{1}}{{2}}{{\left[{e}^{{-{x}^{{2}}+{2}}}\right]}_{{0}}^{{1}}}\)
\(\displaystyle=-\frac{{1}}{{2}}{\left[{e}^{{1}}-{e}^{{2}}\right]}\)
\(\displaystyle=-\frac{{1}}{{2}}{\left[{e}-{e}^{{2}}\right]}=\frac{{1}}{{2}}{\left[{e}^{{2}}-{e}\right]}\)
\(\displaystyle\because{\int_{{0}}^{{1}}}{x}{e}^{{-{x}^{{2}}+{2}}}{\left.{d}{x}\right.}=\frac{{1}}{{2}}{\left({e}^{{2}}-{e}\right)}\)
0

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