A use integration by parts to show that int fx d xx fx int x...

(a) Use integration by parts to show that $$ \int f(x) d x=x f(x)-\int x f^{\prime}(x) d x $$ (b) If $f$ and $g$ are inverse functions and $f^{\prime}$ is continuous, prove that $$ \int_{a}^{b} f(x) d x=b f(b)-a f(a)-\int_{f(a)}^{(j)} g(y) d y $$ [Hint: Use part (a) and make the substitution $y=f(x) . ]$ (c) In the case where $f$ and $g$ are positive functions and $b>a>0,$ draw a diagram to give a geometric interpretation of part (b). (d) Use part (b) to cvaluate $\int_{1}^{c} \ln x d x$ .

Transcript

00:01 This is a great problem from your textbook because this is allowing us to be introduced to the concept of a proof.

00:10 We have to use our knowledge of integral calculus to make conclusions in order to prove something that we're given.

00:18 And this is going to come up very often in future math classes if that's what you're doing.

00:23 But this also helps us build our intuition behind the integration technique that we call integration by.

00:30 Parts.

00:31 So let's get started with our proof.

00:35 So for part a, we're given that we have the integral of f of x in dx, and we have to show that we get a specific integral that we're given by integration by parts.

00:47 So we're going to let you be f of x.

00:49 So that means d u would be f prime of x.

00:52 And then dv would be d u, so v equals x.

00:58 So then let's plug everything in into our integral.

01:01 So the integral of f of x and dx would be equal to uv minus the integral of v d u.

01:08 So we would have f of x times x minus the integral of f, pardon me, of x times f prime of x in dx.

01:18 We're just making these substitutions that we had here before.

01:22 We can simplify that to say that we get x times f of x minus the integral of x times f prime of x in dx.

01:32 So that means we proved what we wanted to show.

01:36 Now on to part b, we have to take this one step further.

01:39 We're asked, well, what's going to happen if i put limits of integration into this integral? what's going to happen? prove this result.

01:49 So that's what we're going to do.

01:51 We're going to have the integral from a to b of f of x in dx.

01:56 Well, that means we would take what we have before and evaluate it at those limits.

02:01 So we would have x times f of x minus the integral of x times f prime of x in dx evaluated from a to b.

02:12 So let's plug everything in.

02:14 We would get, we separate these parts because remember we can separate these parts based on the properties of integrals.

02:21 We can separate this to say we have x times f of x evaluated from a to b minus the integral of x f prime of x, in dx evaluated from a to b...

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