Question

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

Answer 1

`0`

Explanation:

We have that: `int_-2^2 5x^(1/3)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 5x^(1/3)dx=[5*3/4x^(4/3)]_-2^2`

`=[15/4x^(4/3)]_-2^2`

Now evaluate by simply substituting in the limits like so:

`={15/4(2)^(4/3)}-{15/4(-2)^(4/3)}`

`={15/4(16)^(1/3)}-{15/4(16)^(1/3)}`

`=0`

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

`-> 5(-x)^(1/3)=-5x^(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]}