Lesson Video: Integration by Substitution: Definite Integrals Mathematics

Video Transcript

In this lesson, we’ll learn how to use integration by substitution to evaluate definite integrals. At this stage, you should feel comfortable finding the antiderivative for a variety of functions, including polynomials, trigonometric, and logarithmic functions. In this lesson we’ll look at how to apply these rules to find the antiderivative for more complicated functions.

Because of the fundamental theorem of calculus, it’s important to be able to find the antiderivative, but our formulae don’t tell us how to evaluate integrals such as the definite integral between one and two of 𝑥 squared times 𝑥 cubed minus three cubed. So, to find this integral, we use a special strategy of introducing something extra, a new variable. This is called integration by substitution. And it’s sometimes referred to as the reverse chain rule.

The first step is always to get our integral into this form. It’s the integral of 𝑓 of 𝑔 of 𝑥 times 𝑔 prime of 𝑥 with respect to 𝑥. Notice that we have 𝑔 of 𝑥 as the inside part of a composite function and then its derivative 𝑔 prime of 𝑥. Then, once we’re sure that our integral is in this form, we apply the substitution rule for definite integrals.

This says that if 𝑔 prime is continuous on the closed interval 𝑎 to 𝑏, and 𝑓 is continuous on the range of 𝑢 which is 𝑔 of 𝑥, then the definite integral between 𝑎 and 𝑏 of 𝑓 of 𝑔 of 𝑥 times 𝑔 prime of 𝑥 with respect to 𝑥 is equal to the definite integral between 𝑔 of 𝑎 and 𝑔 of 𝑏 of 𝑓 of 𝑢 with respect to 𝑢. It’s often best just to have a look at an example of how this works.

Evaluate the definite integral between one and two of 𝑥 squared times 𝑥 cubed minus three cubed with respect to 𝑥.

This is not a polynomial that’s nice to integrate using our standard rules of finding the antiderivative. And we certainly don’t want to distribute our parentheses and find the antiderivative for each term. Instead, we spot that our integral is set up in this form. It’s the definite integral between some limits of 𝑎 and 𝑏 of some function of 𝑔 of 𝑥 times the derivative of the inner part of that composite function. Here, 𝑔 of 𝑥, the inner part of our composite function, is 𝑥 cubed minus three.

And then, we have a scalar multiple of its derivative here. So, we use integration by substitution to evaluate the integral. This says that if 𝑔 prime, the derivative of 𝑔, is continuous on some closed interval 𝑎 to 𝑏, and 𝑓 is continuous on the range of 𝑢, which is our function 𝑔 of 𝑥. Then, then the definite integral is equal to the definite integral between 𝑔 of 𝑎 and 𝑔 of 𝑏 of 𝑓 of 𝑢 with respect to 𝑢. So, we let 𝑢 be equal to the function that we defined as 𝑔 of 𝑥. It’s the function inside a function whose derivative also appears.

So, we’re going to let 𝑢 be equal to 𝑥 cubed minus three. Now, this is great as when we differentiate 𝑢 with respect to 𝑥, we see that d𝑢 by d𝑥 equals three 𝑥 squared. And in integration by substitution, we think of d𝑢 and d𝑥 as differentials. And we can alternatively write this as d𝑢 equals three 𝑥 squared d𝑥. Notice that whilst d𝑢 by d𝑥 is definitely not a fraction, we do treat it a little like one in this process. We divide both sides by three. And we see that a third d𝑢 equals 𝑥 squared d𝑥.

So, let’s look back to our original integral. We now see that we can replace 𝑥 squared d𝑥 with a third d𝑢. And we can replace 𝑥 cubed minus three with 𝑢. But what do we do with our limits of one and two? Well, we need to replace them with 𝑔 of one and 𝑔 of two. Well, we go back to our original substitution. We said that 𝑢 is equal to 𝑥 cubed minus three, and our lower limit is when 𝑥 is equal to one. So, that’s when 𝑢 is equal to one cubed minus three, which is equal to negative two.

Our upper limit is when 𝑥 is equal to two. So, at this stage, 𝑢 is equal to two cubed minus three, which is of course five. We can now replace each part of our integral with the various substitution. And we see that we’re going to need to work out the definite integral between negative two and five of a third 𝑢 cubed with respect to 𝑢. Remember, we can take any constant factors outside of the integral and focus on integrating 𝑢 cubed.

And then, we recall that we can integrate a polynomial term whose exponent is not equal to negative one by adding one to that exponent and then dividing by that value. So, the integral of 𝑢 cubed is 𝑢 to the fourth power divided by four. And this means our integral is equal to a third times 𝑢 to the fourth power divided by four evaluated between negative two and five. We’ll substitute 𝑢 equals five and 𝑢 equals negative two, and find their difference.

In this case, that’s simply a third of five to the fourth power divided by four minus negative two to the fourth power divided by four. That gives us a third of 609 over four. And we can cancel through by three. And we obtain our solution to be equal to 203 over four, or 50.75. And so, the definite integral between one and two of 𝑥 squared times 𝑥 cubed minus three cubed with respect to 𝑥 is 50.75.

In this example, we saw that we should try and choose 𝑢 to be some factor of the integrand whose differential also occurs, albeit some scalar multiple of it. If that’s not possible though, we try choosing 𝑢 to be a more complicated part of the integrand. This might be the inner function in a composite function or similar.

Let’s have a look at an example of this form.

Find the definite integral between negative one and four of 𝑥 times the square root of 𝑥 plus five with respect to 𝑥 to the nearest thousandth.

In this question, our integrand is the product of two functions, one of which is itself a composite function. It certainly is a tricky one to integrate. And really, we have two options. We could try the substitution method or integration by parts. Notice though that the inner part of our composite function has a nice simple derivative. And that tells us we might be able to use integration by substitution.

The substitution rule for definite integrals says that if 𝑔 prime is continuous on the closed interval 𝑎 to 𝑏, and 𝑓 is continuous over the range of 𝑢, which is equal to 𝑔 of 𝑥. Then the definite integral between 𝑎 and 𝑏 of 𝑓 of 𝑔 of 𝑥 times 𝑔 prime of 𝑥 with respect to 𝑥 is equal to the definite integral between 𝑔 of 𝑎 and 𝑔 of 𝑏 of 𝑓 of 𝑢 with respect to 𝑢. So, with this method, we try to choose 𝑢, which is 𝑔 of 𝑥, to be some factor of the integrand whose differential also occurs, albeit a scalar multiple of it. Here though, it’s not instantly obvious what that might be.

So, instead, we choose 𝑢 to be some more complicated part of the function, here, the inner part in a composite function that has a nice derivative. We’ll try 𝑢 equals 𝑥 plus five. The derivative of 𝑥 plus five with respect to 𝑥 is simply one. And whilst we know that d𝑢 by d𝑥 isn’t a fraction, we do treat it a little like one. We treat d𝑢 and d𝑥 as differentials. And we can say that d𝑢 equals d𝑥. Now, this may not seem instantly helpful, as if we replace d𝑥 with d𝑢 and 𝑥 plus five with 𝑢, we still have a part of our function that’s in terms of 𝑥.

However, if we look back to our substitution, we can rearrange this and say that 𝑥 must be equal to 𝑢 minus five. So, we can now replace each part of our integrand and d𝑥 with d𝑢. But what about our limits? Well, here we use our substitution to redefine these. Our lower limit is when 𝑥 is equal to negative one. And since 𝑢 was equal to 𝑥 plus five, 𝑢 is equal to negative one plus five, which is equal to four. Then, when 𝑥 is equal to four, that’s our upper limit, 𝑢 is equal to four plus five, which is nine.

So, we rewrite our definite integral as the definite integral between four and nine, those are our new limits, of 𝑢 minus five — remember, we said 𝑥 is equal to 𝑢 minus five — times the square root of 𝑢 with respect to 𝑢. And actually, let’s rewrite the square root of 𝑢 as 𝑢 to the power of one-half. And then, we can distribute the parentheses. When we multiply 𝑢 by 𝑢 to the power of one-half, we add their exponents. And we end up with 𝑢 to the power of three over two. So, our integrand is 𝑢 to the power of three over two minus five 𝑢 to the power of one-half.

When integrating simple polynomial terms whose exponent is not equal to negative one, we add one to the exponent and divide by this new value. So, when we integrate 𝑢 to the power of three over two, we get 𝑢 to the power of five over two divided by five over two. And that’s the same as two-fifths times 𝑢 to the power of five over two. Similarly, when we integrate negative five 𝑢 to the power of one-half, we get negative five 𝑢 to the power of three over two divided by three over two. And that simplifies to negative 10 over three 𝑢 to the power of three over two.

Now, of course, we mustn’t forget that we’re going to need to evaluate this between the limits of four and nine. That’s two-fifths of nine to the power five over two minus ten-thirds of nine to the power of three over two minus two-fifths times four to the power of five over two minus ten-thirds times four to the power of three over two. That’s 316 over 15, which correct to the nearest thousandth is 21.067.

In our next example, we’ll consider how this process also works for fractional functions.

Determine the definite integral between negative five and negative two of two over the square root of 𝑥 plus six d𝑥.

It may not be instantly obvious how we’re going to evaluate this integral. However, if we look carefully, we can see that the numerator is a scalar multiple of the derivative of the inner function on our denominator. In other words, the derivative of 𝑥 plus six multiplied by two is equal to the numerator. That’s two. And that’s a hint to us that we’re going to need to use integration by substitution to evaluate our integral.

Remember, in integration by substitution, we introduce a new function. That is 𝑢, which we let 𝑏 equal to 𝑔 of 𝑥. And we see that the definite integral between 𝑎 and 𝑏 of 𝑓 of 𝑔 of 𝑥 times 𝑔 prime of 𝑥 with respect to 𝑥 is equal to the definite integral between 𝑔 of 𝑎 and 𝑔 of 𝑏 of 𝑓 of 𝑢 with respect to 𝑢. We’re going to let 𝑢 be equal to the inner part of our composite function, that’s 𝑥 plus six, so that d𝑢 by d𝑥 is equal to one. And whilst d𝑢 by d𝑥 isn’t a fraction, we are allowed to treat it a little like one when we’re working with integration by substitution. And we can say that d𝑢 is equal to d𝑥.

And so, we can replace d𝑥 with d𝑢 and 𝑥 plus six with 𝑢. But we’re going to need to do something with our limits. We use our substitution to redefine them. Our lower limit is when 𝑥 is equal to negative five. So, 𝑢 then is equal to 𝑥 plus six, which is here negative five plus six, which is of course one. Then, our upper limit is when 𝑥 is equal to negative two. So, 𝑢 is equal to negative two plus six, which is four. And so, our integral is equal to the definite integral between one and four of two over the square root of 𝑢 with respect to 𝑢.

Now, in fact, we can write one over the square root of 𝑢 as 𝑢 to the power of negative one-half. And this next step isn’t entirely necessary, but it can make the process simpler. We recall that we can take any constant factors outside of the integral and focus on integrating the function in 𝑢 itself. So, this is equal to two times the definite integral between one and four of 𝑢 to the power of negative one-half. Now, when we integrate 𝑢 to the power of negative one-half, we add one to the exponent and then divide by that new number. So, 𝑢 to the power of negative half becomes 𝑢 to the power of one-half divided by one-half, which is the same as two times 𝑢 to the power of one-half.

We then replace 𝑢 with four and one and find the difference. We get two times two lots of four to the power of one-half minus two lots of one to the power of one-half. Well, four to the power of one-half is two. And one to the power of one-half is one. So, we have two times two times two minus two times one, which is simply equal to four. And we’re done. Since we changed our limits, we don’t need to do anything more. Our definite integral is equal to four.

In the previous two examples, we saw how we can perform integration by substitution, even if it’s not instantly obvious what it might look like. We’re now going to see how we can use the process to integrate a more complicated trigonometric function.

Find the definite integral between zero and 𝜋 by four of negative nine tan 𝑧 times sec squared 𝑧 with respect to 𝑧.

Firstly, let’s not worry that this function is in terms of 𝑧. We’re integrating with respect to 𝑧, so we perform the process as normal. Then, we notice that sec squared 𝑧 is the derivative of tan of 𝑧. This tells us we can use integration by substitution to evaluate it. We’re going to let 𝑢 be equal to tan of 𝑧. And we know that the first derivative of tan of 𝑧 is sec squared 𝑧. And whilst d𝑢 by d𝑧 is not a fraction, we treat it a little like one, and we say that d𝑢 is equal to sec squared 𝑧 d𝑧.

And we now see that we can replace tan of 𝑧 with 𝑢 and we can replace sec squared 𝑧 d𝑧 with d𝑢. Before we go any further though, we’re going to need to work out what our new limits are. So, we use our substitution. We said 𝑢 is equal to tan of 𝑧 and our lower limit is when 𝑧 is equal to zero, which is when 𝑢 is equal to tan of zero, which is zero. Our upper limit is 𝑧 is equal to 𝜋 by four. So, 𝑢 is equal to tan of 𝜋 by four, which is one. And that’s great because we can now rewrite our definite integral as the definite integral between zero and one of negative nine 𝑢 with respect to 𝑢.

Now, we could, if we wanted, take negative nine out as a constant factor. Of course, we don’t need to. And if we don’t, we find that the integral of negative nine 𝑢 is negative nine 𝑢 squared divided by two. We’re going to evaluate this between zero and one. That’s negative nine times one squared over two minus negative nine times zero squared over two, which is just simply negative nine over two. Our definite integral is equal to negative nine over two.

In our very final example, we’ll look at how to use integration by substitution to evaluate the integral of a logarithmic function.

Determine the definite integral between one and 𝑒 of the natural log of 𝑥 over 𝑥 with respect to 𝑥.

In order to evaluate this integral, we need to spot that the derivative of the natural log of 𝑥 is one over 𝑥 and that a part of this function is indeed a scalar multiple of one over 𝑥. We, therefore, let 𝑢 be equal to the natural log of 𝑥. And when we differentiate 𝑢 with respect to 𝑥, we get one over 𝑥. Now, d𝑢 by d𝑥 is not a fraction, but we treat it a little like one. And we see that this is equivalent to saying d𝑢 equals one over 𝑥 d𝑥. And that’s great because we can now replace the natural log of 𝑥 with 𝑢 and one over 𝑥 d𝑥 with d𝑢.

We are going to need to change our limits. So, we use our substitution. We said that 𝑢 is equal to the natural log of 𝑥. And when 𝑥 is equal to one, 𝑢 must be equal to the natural log of one, which is zero. And when 𝑥 is equal to 𝑒, that’s the upper limit, 𝑢 is equal to the natural log of 𝑒, which is one. And that’s great because our definite integral is now equal to the definite integral between zero and one of 𝑢 with respect to 𝑢.

The integral of 𝑢 is 𝑢 squared over two. And when we evaluate this between the limits of zero and one, we get one squared over two minus zero squared over two, which is of course one-half. The definite integral between one and 𝑒 of the natural log of 𝑥 over 𝑥 with respect to 𝑥 is a half.

In this video, we saw that the substitution rule for definite intervals says that if 𝑔 prime, the first derivative of 𝑔, is continuous on the closed interval of 𝑎 to 𝑏, and 𝑓 itself is continuous on the range of 𝑢, which is equal to 𝑔 of 𝑥. Then the definite integral between 𝑎 and 𝑏 of 𝑓 of 𝑔 of 𝑥 times 𝑔 prime of 𝑥 with respect to 𝑥 is equal to the definite integral between 𝑔 of 𝑎 and 𝑔 of 𝑏 of 𝑓 of 𝑢 with respect to 𝑢.

We saw that we usually try to choose 𝑢 to be some factor of the integrand whose differential also occurs, albeit some scalar multiple of it. If that’s not instantly obvious though, we try choosing 𝑢 to be some more complicated part of the integrand. It might be the inner function in a composite function. And we saw that it’s really important that we use our substitution to change our limits before performing the integral. And we saw this method can be used to integrate functions involving roots, trigonometric functions, and logarithms.