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Math 104: Calculus16 chapters | 135 lessons | 11 flashcard sets

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Lesson Transcript

Instructor:
*Kelly Sjol*

Some integrals are as easy as riding a bike. But more often, integrals can look like deformed bikes from Mars in the year 3000. In this lesson, you will learn how to transform these scary-looking integrals into simpler ones that really are as easy as riding a bike.

When you're trying to find a definite or indefinite integral, things may not be particularly clear, and you can't just look it up in a table all the time. In order to find the integral, or the area under the curve, you need a few more tools.

The first tool is the **chain rule**. Remember that if we have a derivative of a composite function like *d*/*dx*(*f(g(x)*)), we sometimes write it as *d*/*dx*(*f(u)*), where *u* is *g(x)*, so *u* is a function of *x*. Then its derivative becomes *f`u* * *du*/*dx* , which equals *f* `(*u*)*u*`. So we're taking the derivative of the outer function and multiplying it times the derivative of the inner function.

We also know that you use the chain rule for derivatives whenever you see parentheses, as well as for **composite functions**. A composite function is like when you take some meat and you put it into a function to make a patty. Then you put that patty into a function to make a burger. So you put the function for making a patty into the function for making a burger. You put meat in, and you get out a hamburger. How does this relate to integration?

Well, if by using the chain rule you write *d*/*dx* * *f(u)*= *f*`(*u*)*u*` and you integrate both sides, you end up with the integral of (*f*`(*u*)*u*`)*dx* = *f(u)*. So this equals our original function. Hmm, what does this mean? This means you can solve by something called substitution.

We're going to substitute something for the inner function in our composite function. Let me explain. You're going to take an integral *f*(*g(x)*), which is a composite function, times *g`(x)* (the derivative of the inside, the 'patty-forming,' function), times *dx*. That equals the anti-derivative of your function evaluated at *g(x)*, or *F*(*g(x)*). And if this is an indefinite integral, you're going to add a constant of integration, *C*.

So how do you do this? First, you're going to make what's called a **substitution**, and you're going to say there's some function, *u*, that is equal to *g(x)*. Then you're going to take the derivative of *u* - that's *du* - which is equal to *g`(x)dx*. Once you've made the substitution, you're going to find the anti-derivative of your function *f*. Finally, once you have the anti-derivative, you're going to substitute *u* back into your function and that should give you the integral of your original function. Make sure by checking it. So what does all of this mean? Let's do an example.

Let's say we have the integral of 3*e*^3*x* * *dx*. There are some implied parentheses here, around the 3*x*. So I'm going to substitute my variable *u* for 3*x*, *u*=3*x*. Then I'm going to take the derivative of *u*, differentiating this with respect to *x*, and write *du*=3*dx*. Then I'm going to use these two substitutions in my original integral. So I have the integral 3*e*^(3*x*) *dx*, which I can rewrite as the integral of *e*^(3*x*) 3*dx*. Well 3*dx* is equal to *du*, and *e*^(3*x*) is the same as *e*^*u*, because *u*=3*x*; that's how I defined it.

So my new function is *e*^*u*, and I'm integrating with respect to *u* now. Let's actually do that integral. The integral of *e*^*u* * *du* is just *e*^*u* + some constant *C*. We have our constant because this is an indefinite integral, and we have *e*^*u* because I know that the derivative of *e*^*u* is *e*^*u* and the integral of *e*^*u* is *e*^*u*. It's that function that never changes. Okay, so now I have everything in terms of *u*. Let's get rid of the *u* by substituting *u*=3*x* into this equation, and I get *e*^(3*x*) + *C*. By using what's called *u* substitution, I have solved the integral of 3*e*^(3*x*) *dx* as *e*^(3*x*) + *C*. But before I go ahead and say this is right, let's check the answer.

Let's take the derivative of *e*^(3*x*) + *C*. So *d*/*dx*(*e*^(3*x*) + *C*) is equal to the derivative of *e*^(3*x*) + the derivative of *C*. The derivative of a constant is 0. For the derivative of *e*^(3*x*) I have to use the chain rule. So I get the derivative of the outside, *e*^(3*x*), times the derivative of the inside, *d*/*dx*(3*x*). If I take that derivative, I get 3, and I end up with 3*e*^(3*x*). What I got by taking the derivative of the right-hand side equals my function on the left-hand side, inside the integral, my integrand. That's good! If it wasn't the same function, that would mean I screwed up somewhere. This is a good sign.

Let's try another one. Let's try the integral of sin(3*x*). So I've got parentheses around 3*x*. Let's substitute *u*, so *u*=3*x*. If *u*=3*x*, then the derivative of *u*, which I'm going to call *du*, is equal to 3*dx*. Now I don't have a 3*dx* in my original equation to substitute, so I'm going to solve the equation for *dx* by dividing both sides by 3. I get *dx*=1/3(*du*). Let's plug that in for *dx* in my original equation, and let's plug in *u* for 3*x*. Now I can integrate this. It's just 1/3 the integral of sin(*u*)*du*. The integral of the sine of something is just minus cosine, so this integral becomes 1/3(-cos(*u*)) + *C*. If I plug in what I had for *u* so I get rid of all of my *u*s, I end up with 1/3(-cos(3*x*)) + *C* - because this is an indefinite integral I'm going to make sure to add the constant at the end. And I have solved the integral: sin(3*x*)*dx*=-1/3(cos(3*x*)) + *C*.

Let's check it. Let's take the derivative of -1/3(cos(3*x*)) + *C*. I get -1/3 * *d*/*dx*(cos(3*x*)) + 0, because the derivative of a constant is 0. I'm going to have to use the chain rule, and I get -1/3(-sin(3*x*)), which is the derivative of the outside function, times the derivative, *d*/*dx*, of 3*x*. The derivative of 3*x* is just 3, so I end up with, cancelling the 3s, sin(3*x*). That matches my initial integrand, so it looks like I didn't screw up.

For these smaller substitutions, all we did was take the **chain rule** in reverse. We substituted *u* for our inner function, and wrote *du* as the derivative of *u* times *dx*. So *u*=*g(x)*, that's our inner function, and *du* is *g*`*dx*. Then we found the anti-derivative of our new function that depended on *u*, *f(u)*. Once we found that, we plugged *u*=*g(x)* back into our anti-derivative to find the anti-derivative of our original function with respect to *x*. Finally, we checked everything to make sure that if we took the derivative of our result, we ended up with the integrand of our original integral.

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Math 104: Calculus16 chapters | 135 lessons | 11 flashcard sets

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