# How do you evaluate the definite integral int(5x^(1/3))dx from [-2,2]?

Jan 10, 2018

$0$

#### Explanation:

We have that: ${\int}_{-} {2}^{2} 5 {x}^{\frac{1}{3}} \mathrm{dx}$

where the limits applied to integral come from the interval you have been asked to evaluate. To begin simply integrate the function:

${\int}_{-} {2}^{2} 5 {x}^{\frac{1}{3}} \mathrm{dx} = {\left[5 \cdot \frac{3}{4} {x}^{\frac{4}{3}}\right]}_{-} {2}^{2}$

$= {\left[\frac{15}{4} {x}^{\frac{4}{3}}\right]}_{-} {2}^{2}$

Now evaluate by simply substituting in the limits like so:

$= \left\{\frac{15}{4} {\left(2\right)}^{\frac{4}{3}}\right\} - \left\{\frac{15}{4} {\left(- 2\right)}^{\frac{4}{3}}\right\}$

$= \left\{\frac{15}{4} {\left(16\right)}^{\frac{1}{3}}\right\} - \left\{\frac{15}{4} {\left(16\right)}^{\frac{1}{3}}\right\}$

$= 0$

It is also possible to arrive at this result intuitively by exploiting the symmetry of the function.

$\to 5 {\left(- x\right)}^{\frac{1}{3}} = - 5 {x}^{\frac{1}{3}}$

That is, the function has odd symmetry. If we plot this function we can see clearly that the function is reflected but negative through the y -axis. As a result, over the interval the area above the x-axis will exactly cancel with the area under the x-axis giving us $0$.

graph{5x^(1/3) [-18.19, 18.19, -9.1, 9.09]}