0:00
All right.
00:01
We have a couple problems to do, and they're pretty much the same.
00:04
We have the integral from 0 to 3 of the square root of y plus 1, d .y.
00:11
And then the second problem, they just change the bounds from 0 to 3 to be from negative 1 to 0 of the square root of y plus 1, d .y.
00:25
So no matter what in either of these problems, you're going to let u equal y plus 1, because then du will equal d -y and when you go to rewrite that problem, i can replace this y plus one with u.
00:41
And instead of writing the square root, i'm going to write it to the one -half power, and d -y is equal to d -u.
00:48
The difference, though, is when you go to substitute zero in for y, because you're going from in terms of y than in terms of you, zero plus one would give me one, and then again, and plug in 3.
01:02
3 plus 1 gives me 4.
01:04
So again, really the same work on this other problem for letter b.
01:08
It's u to the 1 half power, du.
01:11
But as you plug in your bounds, negative 1 plus 1 and be 0, and then plug in 0 plus 1 and be 1.
01:17
So then your next step would be to find the antiderivative of that.
01:21
Add 1 to your exponent and multiply by the reciprocal...