# Evaluate the integral. \int_{-2}^{2}(x+3)^{2}dx

Evaluate the integral.
$$\displaystyle{\int_{{-{2}}}^{{{2}}}}{\left({x}+{3}\right)}^{{{2}}}{\left.{d}{x}\right.}$$

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Robert Harris
Step 1
Given:
The integral
$$\displaystyle{\int_{{-{2}}}^{{{2}}}}{\left({x}+{3}\right)}^{{{2}}}{\left.{d}{x}\right.}$$
Step 2
Use substitution to integrate
Let u=x+3
$$\displaystyle\Rightarrow{d}{u}={\left.{d}{x}\right.}$$
When x = – 2 , u = – 2 + 3 = 1
When x = 2 , u = 2 + 3 = 5
Substituting u=x+3, we get the integral
$$\displaystyle{\int_{{{1}}}^{{{5}}}}{u}^{{{2}}}{d}{u}$$
Step 3
To simplify further, use the power rule of integration
$$\displaystyle\int{x}^{{{n}}}{\left.{d}{x}\right.}={\frac{{{x}^{{{n}+{1}}}}}{{{\left({n}+{1}\right)}}}}+{C}$$
$$\displaystyle{\int_{{{1}}}^{{{5}}}}{u}^{{{2}}}{d}{u}={{\left[{\frac{{{u}^{{{3}}}}}{{{3}}}}\right]}_{{{1}}}^{{{5}}}}$$
$$\displaystyle={\left({\frac{{{5}^{{{3}}}}}{{{3}}}}\right)}-{\left({\frac{{{1}}}{{{3}}}}\right)}$$
$$\displaystyle={\frac{{{125}}}{{{3}}}}-{\frac{{{1}}}{{{3}}}}$$
$$\displaystyle={\frac{{{125}-{1}}}{{{3}}}}$$
$$\displaystyle={\frac{{{124}}}{{{3}}}}$$
Therefore,
$$\displaystyle{\int_{{-{2}}}^{{{2}}}}{\left({x}+{3}\right)}^{{{2}}}{\left.{d}{x}\right.}={\frac{{{124}}}{{{3}}}}$$
###### Have a similar question?
Susan Yang
Step 1: If f(x) is a continuous function from a to b, and if F(x) is its integral, then:
$$\displaystyle{\int_{{{a}}}^{{{b}}}}{f{{\left({x}\right)}}}{\left.{d}{x}\right.}={F}{\left({x}\right)}{{\mid}_{{{a}}}^{{{b}}}}={F}{\left({b}\right)}-{F}{\left({a}\right)}$$
Step 2: In this case, $$\displaystyle{f{{\left({x}\right)}}}={\left({x}+{3}\right)}^{{{2}}}$$. Find its integral.
$$\displaystyle{\frac{{{x}^{{{3}}}}}{{{3}}}}+{3}{x}^{{{2}}}+{9}{x}{{\mid}_{{-{2}}}^{{{2}}}}$$
Step 3: Since $$\displaystyle{F}{\left({x}\right)}{{\mid}_{{{a}}}^{{{b}}}}={F}{\left({b}\right)}-{F}{\left({a}\right)}$$, expand the above into F(2)−F(−2):
$$\displaystyle{\left({\frac{{{2}^{{{3}}}}}{{{3}}}}+{3}\times{2}^{{{2}}}+{9}\times{2}\right)}-{\left({\frac{{{\left(-{2}\right)}^{{{3}}}}}{{{3}}}}+{3}{\left(-{2}\right)}^{{{2}}}+{9}\times-{2}\right)}$$
Step 4: Simplify.
$$\displaystyle{\frac{{{124}}}{{{3}}}}$$