(a) Use integration by parts to show that ∫f(x) d x=x...

Key Concepts

Geometric Interpretation of Definite Integrals

Definite integrals often have a geometric interpretation as the area under a curve, and this perspective can be extended to understand the relationship between a function and its inverse. When a function and its inverse are both positive, visualizing the integral areas helps in understanding symmetry properties and the duality between the original function's graph and its inverse, which can be represented by reflecting the graph about the line y = x.

Substitution in Integration

The substitution method is a powerful tool in integration used to simplify an integral by changing variables. By substituting a new variable in place of an expression, the integration process becomes more manageable. This technique is particularly useful when dealing with composite functions or transforming an integral involving inverse functions, as it aligns the integrals with functions expressed in terms of their natural variables.

Inverse Functions and Their Properties

Inverse functions reverse the roles of the input and output of a function, and their properties are crucial when establishing relationships between integrals involving a function and its inverse. An important aspect is that the derivative of an inverse function can be expressed in terms of the derivative of the original function, a relationship that is useful in several areas of calculus, particularly when interchanging variables during integration.

Integration by Parts

Integration by parts is a fundamental technique in calculus used to integrate products of functions. It is based on the product rule for differentiation and allows one to transform a difficult integral into another form that is often easier to evaluate. This method splits the integrand into two parts and strategically differentiates one while integrating the other.

Transcript

00:01 Okay, all right, so first they want us to show this part a.

00:18 Okay, if there's not part a, what is i can see? times f prime of x minus the integral.

00:28 Oh, sorry, sorry.

00:33 Of x, prime of x, dx.

00:39 Okay, sorry about that.

00:43 Okay, whatever.

00:44 It's very long.

00:45 Anyways, all right, so they want to use this to integrate my parts, and remember here's the formula for.

00:51 Which is my heart parts.

00:54 And scroll of udv is equal to uv minus its scroll of v.

01:01 Du.

01:01 All right, so we can let u equal f of x, and then let dv equal d x and v .o b x and d uv f prime of x d x, right? now using the formula, uv, which be x f of x and then v d u will just be x times f prime of x all right there you now that's part a now you see how we use the third parts and we use g you good at f of x and um dv equal d x and we arrived at the conclusion um that's part a of part b so just do this right part b um if f and g or inverse functions and f prime is continuous prove that are this and ff b f of a g of y d y and all right so we can we can look at their hint and the hint says use use part a mix substitution y equals on f x okay so let's let's do that all right so first let's just uh expand this out using insurgit from my parts and see what we would get um well using uh what we have here.

03:12 We'll see that there will be an f, i mean, sorry, an x, f of x, but this will be evaluated from b to a.

03:22 So that's important to note, i guess i should have.

03:27 It should be evaluated from b to a.

03:31 And then the other part will be b to a of x, f prime of x, d x.

03:43 Now they said make the substitution y, equals f x.

03:46 Okay.

03:47 So first, let's just expand this out, this part out.

03:52 And we see that this gives us just bf of b minus a, f of a, which does correspond to this part.

04:01 Right.

04:01 So these parts, those are equal.

04:02 But now we have to prove that integral of b to a of x prime of x, dx is equal to integral of f of b to f of a of g of y d y and remember um according to this f and g are inverses it's something to know um so now let's let's um it'll be you you see what i mean right so so let's let's let's let y equal of fx actually let me add to it on side and do like the other side um so we have been in this right.

04:59 So y equals f of x.

05:08 I'm just going to read it again.

05:10 Y equal fx.

05:12 So now, let's see.

05:14 All right.

05:14 So what will we do here? so first, we might notice that if we were to, if we isolate x, we would get, give me a second.

05:31 Okay, so f, okay, so if we, i just, if we said f inverse of both sides, notice that we'll be saying inverse of the function itself, which just gives us x itself.

05:55 And f inverse is, remember, the inverse is just g.

06:04 The inverse of f is just g.

06:05 So we can put g of y there, x is equal to g of y.

06:08 So we can plug that in and you can see that this will give us the end scroll of b to a of g of y.

06:17 So we have the x part.

06:18 Now we do out f prime of x.

06:20 So let's go back here.

06:22 So if we just did f prime of x here, then we will get dy, dx equal to f prime of x.

06:31 Okay, so now we can take that.

06:33 Put that here, dy, dx, and then times the dx here.

06:37 We're just plug in d.

06:38 Y, dx for f prime vx, and take in dx already there.

06:42 And then we see that these dx is canceled out, and we have the integral of, oh, wait, wait, yep, sorry, this is important part to include.

06:52 So now, notice how the intervals will, it won't be bna anymore.

06:56 They will be changing.

06:57 This is because we're subbing in, we're substituting.

07:04 And here, if we want to figure out what those values would be, we just plug them in here.

07:11 And we'll see that would be f of b, and then two f of a.

07:17 So these should be f of b and f of a.

07:26 And so it's f of b, f of a, g, gy, d .y, d.

07:37 So now we'll write this more clearly.

07:43 So we see that the integral from b to a of fx, dx, is initially equal to, using our, using our uh what we did for uh part a this f uh f the x times f of x minus um so just be we can say so f times f x b to a minus integral we see this um and then what they gave us which was integral of um b to right here, b to a of f of x, x, x equal to, okay, just making sure that that...

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