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

1 Answer
Andrew I.
Jan 10, 2018

#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]}

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