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

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

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
$$\displaystyle=-\frac{{1}}{{2}}{{\left[{e}^{{-{x}^{{2}}+{2}}}\right]}_{{0}}^{{1}}}$$
$$\displaystyle=-\frac{{1}}{{2}}{\left[{e}^{{1}}-{e}^{{2}}\right]}$$
$$\displaystyle=-\frac{{1}}{{2}}{\left[{e}-{e}^{{2}}\right]}=\frac{{1}}{{2}}{\left[{e}^{{2}}-{e}\right]}$$
$$\displaystyle\because{\int_{{0}}^{{1}}}{x}{e}^{{-{x}^{{2}}+{2}}}{\left.{d}{x}\right.}=\frac{{1}}{{2}}{\left({e}^{{2}}-{e}\right)}$$

### Relevant Questions

Evaluate the following definite integrals:
$$\displaystyle{\int_{{0}}^{{1}}}{\left({x}^{{4}}+{7}{e}^{{x}}-{3}\right)}{\left.{d}{x}\right.}$$
Use a table of integrals to evaluate the definite integral.
$$\displaystyle{\int_{{0}}^{{3}}}\sqrt{{{x}^{{2}}+{16}}}{\left.{d}{x}\right.}$$
Evaluate the following definite integrals
$$\displaystyle{\int_{{{0}}}^{{{1}}}}{x}{e}^{{{\left(-{x}^{{{2}}}+{2}\right)}}}{\left.{d}{x}\right.}$$
Evaluate the following definite integrals
$$\displaystyle{\int_{{{0}}}^{{{1}}}}{\left({x}^{{{4}}}+{7}{e}^{{{x}}}-{3}\right)}{\left.{d}{x}\right.}$$
Evaluate the integrals
$$\displaystyle{\int_{{0}}^{{1}}}{\left[{t}{e}^{{t}}^{2}+{e}^{{-{{t}}}}{j}+{k}\right]}{\left.{d}{t}\right.}$$
Evaluate the following integrals.
$$\displaystyle{\int_{{-{2}}}^{{-{1}}}}\sqrt{{-{4}{x}-{x}^{{{2}}}}}{\left.{d}{x}\right.}$$
Evaluate the following definite integral.
$$\displaystyle{\int_{{{0}}}^{{{\frac{{\pi}}{{{2}}}}}}}{x}{\cos{{2}}}{x}{\left.{d}{x}\right.}$$
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
$$\displaystyle{\int_{{{1}}}^{{{e}}}}{\ln{{2}}}{x}{\left.{d}{x}\right.}$$
$$\displaystyle\int{\left({2}{x}^{{{3}}}-{x}^{{{2}}}+{3}{x}-{7}\right)}{\left.{d}{x}\right.}$$