How do you evaluate the definite integral int (x^3-pix^2 dx(x3πx2dx from [2,5][2,5]?

1 Answer
Apr 21, 2017

a

Explanation:

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(x3πx2)dx=(x3)dx(πx2)dx
=int(x^3)dx-pi int(x^2)dx=(x3)dxπ(x2)dx
=(x^4/4)-pi(x^3/3)=(x44)π(x33)

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)]_2^552(x3πx2)dx=[(x44)π(x33)]52
=[(5^4/4)-pi(5^3/3)]-[(2^4/4)-pi(2^3/3)]=[(544)π(533)][(244)π(233)]
=[(625/4)-pi(125/3)]-[(16/4)-pi(8/3)]=[(6254)π(1253)][(164)π(83)]
=(625/4)-pi(125/3)-(16/4)+pi(8/3)=(6254)π(1253)(164)+π(83)
=((625-16)/4)-pi((125+8)/3)=(625164)π(125+83)
=609/4-pi133/3=6094π1333
This is exact, but we can approximate if you want
=12.9727257=12.9727257