00:01
For this question, we need to solve the following integral.
00:04
To do this, start by expanding the integrand.
00:07
To expand the integrand, make sure that you multiply every term in the first set of parentheses by every term in the second set of parentheses.
00:16
This gets us a squared minus 2a to the 3 halves x to the 1 half, plus a x minus 2a to the 3 halves, to the one half plus 4 a x minus 2 a to the 1 half x to the 3 halves plus a x minus 2 a to the 1 half x to the 3 halves combine like terms we can combine negative 2 a to the 3 halves x to the 1 half x to the 1 half with negative 2a to the 3 halves x to the 1 half.
01:06
We can also combine negative 2 a to the 1 half x to the 3 halves with the other matching term.
01:13
Finally, we can combine 4ax plus ax plus a x.
01:19
This gets us pi times the integral from 0 to a of a squared minus 4a to the 3 halves x to the 1 half plus 6 a.
01:35
Minus 4 a to the 1 half x to the 3 halves plus x squared d x now take the anti -derivative of the integrand term by term so we're going to leave the factor of pi out front now find the antiderivative of a squared which is just a squared x since a squared is a square is a constant now find the anti -derivative of negative 4, a to the three -haves, x to the one -half.
02:06
We can do this by reversing the power rule.
02:09
This gets us the anti -derivative negative eight -thirds, a to the three - halves, x to the three -haves.
02:18
Now find the anti -derivative of 6a -x.
02:22
This gives us plus 3 a -x squared.
02:26
Now find the anti -derivative of the next term by reversing.
02:30
The power rule.
02:32
This gives us negative eight -fifths, a to the one -half, x to the five -half.
02:40
Finally, find the antiderivative of x squared, which is just one -third x -cubed.
02:47
And this is all evaluated from x equals zero to x equals a.
02:52
Now apply the fundamental theorem of calculus by first plugging in the upper limit of integration a...
