diferira7c

2022-01-03

Use symmetry to evaluate the following integrals.

${\int}_{-2}^{2}({x}^{2}+{x}^{3})dx$

Mason Hall

Beginner2022-01-04Added 36 answers

Step 1

Given: An integral${\int}_{-2}^{2}({x}^{2}+{x}^{3})dx$

By symmetry an integral of form${\int}_{-a}^{a}f\left(x\right)dx$ can be written as ${\int}_{-a}^{0}f\left(x\right)dx+{\int}_{0}^{a}f\left(x\right)dx$

Therefore,${\int}_{-2}^{2}({x}^{2}+{x}^{3})dx$ can be written as:

${\int}_{-2}^{2}({x}^{2}+{x}^{3})dx={\int}_{-2}^{0}({x}^{2}+{x}^{3})dx+{\int}_{0}^{2}({x}^{2}+{x}^{3})dx$

Step 2

Now integrating the above function with respect to x.

${\int}_{-2}^{2}({x}^{2}+{x}^{3})dx={\int}_{-2}^{0}({x}^{2}+{x}^{3})dx+{\int}_{0}^{2}({x}^{2}+{x}^{3})dx$

$={[\frac{{x}^{3}}{3}+\frac{{x}^{4}}{4}]}_{-2}^{0}+{[\frac{{x}^{3}}{3}+\frac{{x}^{4}}{4}]}_{0}^{2}$

$=[(0+0)-(\frac{-8}{3}+4)]+[(\frac{8}{3}+4)-(0+0)]$

$=\frac{8}{3}-4+\frac{8}{3}+4$

$=\frac{16}{3}$

Thus, the value of${\int}_{-2}^{2}({x}^{2}+{x}^{3})dx$ is $\frac{16}{3}$

Given: An integral

By symmetry an integral of form

Therefore,

Step 2

Now integrating the above function with respect to x.

Thus, the value of

Lynne Trussell

Beginner2022-01-05Added 32 answers

Lets

karton

Expert2022-01-11Added 613 answers

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