00:01
In this problem, we're given an indefinite integral, which is the integral of y squared times 2 minus y cubed to the two -thirds power d -y.
00:12
Now, i'm going to note that in the problem statement here, it hasn't been marked which numbers are exponents.
00:20
So this is my guess as to what the integral is, but i think it's this because of how this stuff works out.
00:25
But when you write this stuff down, you really need to mark what are exponents or what are.
00:30
Products.
00:32
So the way that we can go at this is by using substitution of variables.
00:37
So one thing to notice here is we've got an expression 2 minus y cubed and we know that if we differentiate this expression with respect to why, we're going to get a factor of y squared because d y cubed of d y is equal to 3 y squared.
00:54
And we have a y squared here.
00:56
So right away that tells us that we we think we might be able to do some sort of a sub of variables so that this y squared d .y gets absorbed into a differential an element of the substituted variable.
01:10
And i'll show you what i mean right now.
01:12
Let's let z equal 2 minus y cubed.
01:16
So this is going to imply that dz is equal to negative 3 y squared dy.
01:23
So right away we get this y squared dy...