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

Question

Emilia Aguilar · Accepted Answer

#0#

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]}

Adam Adkins · Answer

undefined To evaluate the definite integral $ \int_{-2}^{2} 5x^{1/3} \, dx $:

1. Integrate $ 5x^{1/3} $ with respect to $ x $ to find the antiderivative.
2. Evaluate the antiderivative at the upper and lower limits of integration.
3. Subtract the value of the antiderivative at the lower limit from the value at the upper limit to find the definite integral.

Here are the steps in detail:

1. Integrate $ 5x^{1/3} $ with respect to $ x $:
$$ \int 5x^{1/3} \, dx = \frac{5}{4}x^{4/3} + C $$

2. Evaluate the antiderivative at the upper and lower limits of integration:
$$ \left[\frac{5}{4}x^{4/3}\right]_{-2}^{2} $$

3. Evaluate the antiderivative at $ x = 2 $:
$$ \frac{5}{4}(2)^{4/3} = \frac{5}{4}(2^{4/3}) $$

4. Evaluate the antiderivative at $ x = -2 $:
$$ \frac{5}{4}(-2)^{4/3} = \frac{5}{4}((-2)^{4/3}) $$

5. Subtract the value at the lower limit from the value at the upper limit:
$$ \left[\frac{5}{4}(2^{4/3})\right] - \left[\frac{5}{4}((-2)^{4/3})\right] $$

$$ = \frac{5}{4}(2^{4/3}) - \frac{5}{4}((-2)^{4/3}) $$

$$ = \frac{5}{4}(2^{4/3} - (-2)^{4/3}) $$

$$ = \frac{5}{4}(2^{4/3} - 2^{4/3}) $$

$$ = \frac{5}{4}(0) $$

$$ = 0 $$

So, the value of the definite integral $ \int_{-2}^{2} 5x^{1/3} \, dx $ is $ 0 $.