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

Question
Applications of integrals
asked 2021-02-09
Evaluate the integrals.
\(\displaystyle{\int_{{-{{2}}}}^{{2}}}{\left({3}{x}^{{4}}-{2}{x}+{1}\right)}{\left.{d}{x}\right.}\)

Answers (1)

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,
\(\displaystyle{\int_{{-{{2}}}}^{{2}}}{\left({3}{x}^{{4}}-{2}{x}+{1}\right)}{\left.{d}{x}\right.}={\int_{{2}}^{{23}}}{x}^{{4}}{\left.{d}{x}\right.}-{2}{\int_{{-{{2}}}}^{{2}}}{x}{\left.{d}{x}\right.}+{\int_{{-{{2}}}}^{{21}}}{\left.{d}{x}\right.}\)
\(\displaystyle={3}{{\left[\frac{{x}^{{5}}}{{5}}\right]}_{{-{{2}}}}^{{2}}}-{2}{{\left[\frac{{x}^{{2}}}{{2}}\right]}_{{-{{2}}}}^{{2}}}+{{\left[{x}\right]}_{{-{{2}}}}^{{2}}}\)
\(\displaystyle=\frac{{3}}{{5}}{\left[{2}^{{5}}-{\left(-{2}\right)}^{{5}}\right]}-{\left[{2}^{{2}}-{\left(-{2}\right)}^{{2}}\right]}+{\left[{2}-{\left(-{2}\right)}\right]}\)
\(\displaystyle=\frac{{3}}{{5}}{\left({32}-{\left(-{32}\right)}\right)}-{\left({4}-{4}\right)}+{\left({2}+{2}\right)}\)
\(\displaystyle\frac{{3}}{{5}}{\left({64}\right)}-{0}+{4}\)
\(\displaystyle=\frac{{192}}{{5}}+{4}\)
\(\displaystyle=\frac{{{192}+{20}}}{{5}}\)
\(\displaystyle=\frac{{212}}{{5}}={42.4}\)
0

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