15. Evaluate the following integrals: (a) ∫0^2 (x^2)/(√(x^3...

Transcript

00:01 Okay, so for this problem we are given three different integrals that look very similar, but they may have to be using, we may have to be using some different integration techniques.

00:11 So for the first one, i am looking at this and i see x squared minus one to the one half and an x on the outside.

00:21 So this looks really similar to something like u substitution.

00:25 So i'm going to let my u, oops, i'm going to let my u be everything in the one half.

00:32 So i'm going to let that be x squared minus one.

00:36 Now i want to take the derivative of d u with respect to x, which would get me 2x, and that would get me d u is going to be equal to 2x d x.

00:50 Now i see that i have an x here and i have a dx, but i don't have a 2.

00:56 So that means what i'm going to do is divide out this two.

01:01 On both sides.

01:03 Then that finally gets me d u over 2 is equal to x d x.

01:09 So i can just straight up some substitute in everything that i have.

01:14 So now i get the integral of you then i'm to the one half.

01:21 So that is this term right here and now i have my x d x but i'm going to replace that by du, then it is over two, but i'm just going to pull out the one half because it's a constant so we can just pull that out.

01:36 And now we just use regular integration rule.

01:40 So u to the three halves times two thirds.

01:44 All of this, since we pulled out that constant, we need to multiply it plus c.

01:49 And now we need to substitute back in our u.

01:52 So that gets me x squared minus one, three we have over 3 plus c.

02:02 And this is regular u substitution.

02:05 Now, for this one, we also want to use u substitution, but instead of taking u is equal to just x squared minus 1, i'm going to take this to the, i'm going to take the entire thing as my u.

02:24 Now i'm going to do my d -u equals 1⁄2 times x squared minus 1 to the negative 1 half times 2x.

02:35 And remember that we had a d -x on the bottom on the left side, so right here.

02:41 But i want it over here.

02:43 So i just multiplied.

02:46 That's just a little trick.

02:48 And now let's simplify this down just so it's easier to work with.

02:52 So we have d -u equals x over square root of x squared minus 1, d -x.

03:02 And so i don't have really anything that looks like this in my original integral...

Question

15. Evaluate the following integrals: (a) $\int_0^2 \frac{x^2}{\sqrt{x^3 + 1}} dx$ (b) $\int_0^1 x(x^2 + 1)^3 dx$ (c) $\int_1^5 \frac{x}{\sqrt{2x - 1}} dx$

Question

15. Evaluate the following integrals: (a) $\int_0^2 \frac{x^2}{\sqrt{x^3 + 1}} dx$ (b) $\int_0^1 x(x^2 + 1)^3 dx$ (c) $\int_1^5 \frac{x}{\sqrt{2x - 1}} dx$

Please give Ace some feedback