diferira7c

2022-01-03

Use symmetry to evaluate the following integrals.
${\int }_{-2}^{2}\left({x}^{2}+{x}^{3}\right)dx$

Mason Hall

Step 1
Given: An integral ${\int }_{-2}^{2}\left({x}^{2}+{x}^{3}\right)dx$
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, ${\int }_{-2}^{2}\left({x}^{2}+{x}^{3}\right)dx$ can be written as:
${\int }_{-2}^{2}\left({x}^{2}+{x}^{3}\right)dx={\int }_{-2}^{0}\left({x}^{2}+{x}^{3}\right)dx+{\int }_{0}^{2}\left({x}^{2}+{x}^{3}\right)dx$
Step 2
Now integrating the above function with respect to x.
${\int }_{-2}^{2}\left({x}^{2}+{x}^{3}\right)dx={\int }_{-2}^{0}\left({x}^{2}+{x}^{3}\right)dx+{\int }_{0}^{2}\left({x}^{2}+{x}^{3}\right)dx$
$={\left[\frac{{x}^{3}}{3}+\frac{{x}^{4}}{4}\right]}_{-2}^{0}+{\left[\frac{{x}^{3}}{3}+\frac{{x}^{4}}{4}\right]}_{0}^{2}$
$=\left[\left(0+0\right)-\left(\frac{-8}{3}+4\right)\right]+\left[\left(\frac{8}{3}+4\right)-\left(0+0\right)\right]$
$=\frac{8}{3}-4+\frac{8}{3}+4$
$=\frac{16}{3}$
Thus, the value of ${\int }_{-2}^{2}\left({x}^{2}+{x}^{3}\right)dx$ is $\frac{16}{3}$

Lynne Trussell

$\int \left({x}^{2}+{x}^{3}\right)dx$
Lets

karton

Given:
${\int }_{-2}^{2}\left({x}^{2}+{x}^{3}\right)dx\phantom{\rule{0ex}{0ex}}\int {x}^{2}+{x}^{3}dx\phantom{\rule{0ex}{0ex}}\int {x}^{2}dx+\int {x}^{3}dx\phantom{\rule{0ex}{0ex}}\frac{{x}^{3}}{3}+\frac{{x}^{4}}{4}\phantom{\rule{0ex}{0ex}}\left(\frac{{x}^{3}}{3}+\frac{{x}^{4}}{4}\right){|}_{-2}^{2}\phantom{\rule{0ex}{0ex}}\frac{{2}^{3}}{3}+\frac{{2}^{4}}{4}-\left(\frac{\left(-2{\right)}^{3}}{3}+\frac{\left(-2{\right)}^{4}}{4}\right)\phantom{\rule{0ex}{0ex}}Answer:\phantom{\rule{0ex}{0ex}}\frac{16}{3}$

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