Video: Evaluating the Definite Integral of a Root Function Using Integration by Substitution

Video Transcript

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.