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

Question

Christopher Agresta · Accepted Answer

a

These steps should all be pretty understandable for you, but if not, check this out First, find the indefinite integral #int(x^3-pix^2)dx=int(x^3)dx-int(pix^2)dx# #=int(x^3)dx-pi int(x^2)dx# #=(x^4/4)-pi(x^3/3)# If you didn't understand that last step, this may help. Now that we have the indefinite integral, we can use (one part of) the fundamental theorem of calculus to solve the definite integral. #int_2^5(x^3-pix^2)dx= [(x^4/4)-pi(x^3/3)]_2^5# #=[(5^4/4)-pi(5^3/3)]-[(2^4/4)-pi(2^3/3)]# #=[(625/4)-pi(125/3)]-[(16/4)-pi(8/3)]# #=(625/4)-pi(125/3)-(16/4)+pi(8/3)# #=((625-16)/4)-pi((125+8)/3)# #=609/4-pi133/3# This is exact, but we can approximate if you want #=12.9727257#

Elijah Abram · Answer

undefined To evaluate the definite integral $\int_{2}^{5} (x^3 - \pi x^2) \, dx$, follow these steps:

1. Integrate the function $x^3 - \pi x^2$ with respect to $x$ to find the antiderivative.
2. Evaluate the antiderivative at the upper limit of integration (5).
3. Evaluate the antiderivative at the lower limit of integration (2).
4. Subtract the result from step 3 from the result from step 2.

Let's perform these steps:

1. The antiderivative of $x^3 - \pi x^2$ with respect to $x$ is $\frac{1}{4}x^4 - \frac{\pi}{3}x^3$.
2. Evaluating the antiderivative at $x = 5$, we get $\left(\frac{1}{4}(5)^4 - \frac{\pi}{3}(5)^3\right) = \frac{625}{4} - \frac{125\pi}{3}$.
3. Evaluating the antiderivative at $x = 2$, we get $\left(\frac{1}{4}(2)^4 - \frac{\pi}{3}(2)^3\right) = 4 - \frac{8\pi}{3}$.
4. Subtracting the result from step 3 from the result from step 2, we get $\frac{625}{4} - \frac{125\pi}{3} - (4 - \frac{8\pi}{3})$.
5. Simplifying the expression yields the final result of the definite integral: $\frac{625}{4} - 4 + \frac{125\pi}{3} + \frac{8\pi}{3} = \frac{609}{4} + \frac{133\pi}{3}$.