# Evaluate the following definite integrals \int_{0}^{1}xe^{(-x^{2}+2)}dx

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

2021-02-19
Step 1
To Evaluate the following definite integrals:
Step 2
Given that
$$\displaystyle{\int_{{{0}}}^{{{1}}}}{x}{e}^{{{\left(-{x}^{{{2}}}+{2}\right)}}}{\left.{d}{x}\right.}$$
Let $$\displaystyle{e}^{{{\left(-{x}^{{{2}}}+{2}\right)}}}={u}$$
$$\displaystyle{\frac{{{d}{u}}}{{{\left.{d}{x}\right.}}}}=-{2}{x}{\left({e}^{{-{x}^{{{2}}}+{2}}}\right)}$$
$$\displaystyle{\frac{{{d}{u}}}{{-{2}}}}={x}{e}^{{{\left(-{x}^{{{2}}}+{2}\right)}}}{\left.{d}{x}\right.}$$
Hence $$\displaystyle{\int_{{{0}}}^{{{1}}}}{x}{e}^{{{\left(-{x}^{{-{2}}}+{2}\right)}}}{\left.{d}{x}\right.}={\int_{{{0}}}^{{{1}}}}{\frac{{{d}{u}}}{{-{2}}}}$$
$$\displaystyle{\frac{{-{1}}}{{{2}}}}{{\left[{4}\right]}_{{{0}}}^{{{1}}}}$$
$$\displaystyle={\frac{{-{1}}}{{{2}}}}{\left[{e}^{{{1}}}-{e}^{{{2}}}\right]}$$
$$\displaystyle={\frac{{-{1}}}{{{2}}}}{\left[{e}-{e}^{{{2}}}\right]}={\frac{{{1}}}{{{2}}}}{\left[{e}^{{{2}}}-{e}\right]}$$
$$\displaystyle\because{\int_{{{0}}}^{{{1}}}}{x}{e}^{{{\left(-{x}^{{{2}}}+{2}\right)}}}{\left.{d}{x}\right.}={\frac{{{1}}}{{{2}}}}{\left({e}^{{{2}}}-{e}\right)}$$

### Relevant Questions

Evaluate the following definite integrals
$$\displaystyle{\int_{{{0}}}^{{{1}}}}{\left({x}^{{{4}}}+{7}{e}^{{{x}}}-{3}\right)}{\left.{d}{x}\right.}$$
Evaluate the following definite integral.
$$\displaystyle{\int_{{{0}}}^{{{\frac{{\pi}}{{{2}}}}}}}{x}{\cos{{2}}}{x}{\left.{d}{x}\right.}$$
Evaluate the following definite integrals:
$$\displaystyle{\int_{{0}}^{{1}}}{\left({x}{e}^{{-{x}^{{2}}+{2}}}\right)}{\left.{d}{x}\right.}$$
Evaluate the following integrals.
$$\displaystyle{\int_{{-{2}}}^{{-{1}}}}\sqrt{{-{4}{x}-{x}^{{{2}}}}}{\left.{d}{x}\right.}$$
Evaluate the following definite integrals:
$$\displaystyle{\int_{{0}}^{{1}}}{\left({x}^{{4}}+{7}{e}^{{x}}-{3}\right)}{\left.{d}{x}\right.}$$
Evaluate the following integrals.
$$\displaystyle{\int_{{-{{2}}}}^{{-{{1}}}}}\sqrt{{-{4}{x}-{x}^{{2}}}}{\left.{d}{x}\right.}$$
Which of the following integrals are improper integrals?
1.$$\displaystyle{\int_{{{0}}}^{{{3}}}}{\left({3}-{x}\right)}^{{{2}}}{\left\lbrace{3}\right\rbrace}{\left.{d}{x}\right.}$$
2.$$\displaystyle{\int_{{{1}}}^{{{16}}}}{\frac{{{e}^{{\sqrt{{{x}}}}}}}{{\sqrt{{{x}}}}}}{\left.{d}{x}\right.}$$
3.$$\displaystyle{\int_{{{1}}}^{{\propto}}}{\frac{{{3}}}{{\sqrt{{{3}}}{\left\lbrace{x}\right\rbrace}}}}{\left.{d}{x}\right.}$$
4.$$\displaystyle{\int_{{-{2}}}^{{{2}}}}{3}{\left({x}+{1}\right)}^{{-{1}}}{\left.{d}{x}\right.}$$
a) 1 only
b)1 and 2
c)3 only
d)2 and 3
e)1,3 and 4
f)All of the integrals are improper
$$\displaystyle{\int_{{{\left({1},{1},{2}\right)}}}^{{{\left({3},{5},{0}\right)}}}}{y}{z}{\left.{d}{x}\right.}+{x}{z}{\left.{d}{y}\right.}+{x}{y}{\left.{d}{z}\right.}$$
$$\displaystyle{\int_{{0}}^{{3}}}\sqrt{{{x}^{{2}}+{16}}}{\left.{d}{x}\right.}$$
$$\displaystyle{\int_{{{0}}}^{{\pi}}}{x}{\sin{{x}}}{\left.{d}{x}\right.}$$