# Evaluate the integrals. int_-2^2(3x^4-2x+1)dx

Question
Applications of integrals
Evaluate the integrals.
$$\displaystyle{\int_{{-{{2}}}}^{{2}}}{\left({3}{x}^{{4}}-{2}{x}+{1}\right)}{\left.{d}{x}\right.}$$

2021-02-10
Step 1:Given that
Evaluate integrals.
$$\displaystyle{\int_{{-{{2}}}}^{{2}}}{\left({3}{x}^{{4}}-{2}{x}+{1}\right)}{\left.{d}{x}\right.}$$
Step 2:Solving the integral
We have,
$$\displaystyle{\int_{{-{{2}}}}^{{2}}}{\left({3}{x}^{{4}}-{2}{x}+{1}\right)}{\left.{d}{x}\right.}={\int_{{2}}^{{23}}}{x}^{{4}}{\left.{d}{x}\right.}-{2}{\int_{{-{{2}}}}^{{2}}}{x}{\left.{d}{x}\right.}+{\int_{{-{{2}}}}^{{21}}}{\left.{d}{x}\right.}$$
$$\displaystyle={3}{{\left[\frac{{x}^{{5}}}{{5}}\right]}_{{-{{2}}}}^{{2}}}-{2}{{\left[\frac{{x}^{{2}}}{{2}}\right]}_{{-{{2}}}}^{{2}}}+{{\left[{x}\right]}_{{-{{2}}}}^{{2}}}$$
$$\displaystyle=\frac{{3}}{{5}}{\left[{2}^{{5}}-{\left(-{2}\right)}^{{5}}\right]}-{\left[{2}^{{2}}-{\left(-{2}\right)}^{{2}}\right]}+{\left[{2}-{\left(-{2}\right)}\right]}$$
$$\displaystyle=\frac{{3}}{{5}}{\left({32}-{\left(-{32}\right)}\right)}-{\left({4}-{4}\right)}+{\left({2}+{2}\right)}$$
$$\displaystyle\frac{{3}}{{5}}{\left({64}\right)}-{0}+{4}$$
$$\displaystyle=\frac{{192}}{{5}}+{4}$$
$$\displaystyle=\frac{{{192}+{20}}}{{5}}$$
$$\displaystyle=\frac{{212}}{{5}}={42.4}$$

### Relevant Questions

Evaluate the integrals.
$$\displaystyle{\int_{{-{2}}}^{{{2}}}}{\left({3}{x}^{{{4}}}-{2}{x}+{1}\right)}{\left.{d}{x}\right.}$$
Evaluate the following definite integrals
$$\displaystyle{\int_{{{0}}}^{{{1}}}}{\left({x}^{{{4}}}+{7}{e}^{{{x}}}-{3}\right)}{\left.{d}{x}\right.}$$
Evaluate the following integrals.
$$\displaystyle\int{\left({2}{x}^{{{3}}}-{x}^{{{2}}}+{3}{x}-{7}\right)}{\left.{d}{x}\right.}$$
Evaluate the following integrals.
$$\displaystyle{\int_{{-{2}}}^{{-{1}}}}\sqrt{{-{4}{x}-{x}^{{{2}}}}}{\left.{d}{x}\right.}$$
Evaluate the following definite integrals
$$\displaystyle{\int_{{{0}}}^{{{1}}}}{x}{e}^{{{\left(-{x}^{{{2}}}+{2}\right)}}}{\left.{d}{x}\right.}$$
Show that the differential forms in the integrals are exact. Then evaluate the integrals.
$$\displaystyle{\int_{{{\left({1},{1},{2}\right)}}}^{{{\left({3},{5},{0}\right)}}}}{y}{z}{\left.{d}{x}\right.}+{x}{z}{\left.{d}{y}\right.}+{x}{y}{\left.{d}{z}\right.}$$
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
$$\displaystyle{\int_{{-{{2}}}}^{{-{{1}}}}}\sqrt{{-{4}{x}-{x}^{{2}}}}{\left.{d}{x}\right.}$$
$$\displaystyle{\int_{{{0}}}^{{{\frac{{\pi}}{{{2}}}}}}}{x}{\cos{{2}}}{x}{\left.{d}{x}\right.}$$
$$\displaystyle{\int_{{-{2}}}^{{{2}}}}{\frac{{{d}{p}}}{{{4}-{p}^{{{2}}}}}}$$