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

Find ∫_(βˆ’1) ^(4) π‘₯√(π‘₯ + 5) dπ‘₯ to the nearest thousandth.

03:47

Video Transcript

Find the definite integral between negative one and four of π‘₯ times the square root of π‘₯ plus five with respect to π‘₯ 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 π‘Ž to 𝑏, and 𝑓 is continuous over the range of 𝑒, which is equal to 𝑔 of π‘₯. Then the definite integral between π‘Ž and 𝑏 of 𝑓 of 𝑔 of π‘₯ times 𝑔 prime of π‘₯ with respect to π‘₯ is equal to the definite integral between 𝑔 of π‘Ž and 𝑔 of 𝑏 of 𝑓 of 𝑒 with respect to 𝑒. So, with this method, we try to choose 𝑒, which is 𝑔 of π‘₯, 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 𝑒 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 𝑒 equals π‘₯ plus five. The derivative of π‘₯ plus five with respect to π‘₯ is simply one. And whilst we know that d𝑒 by dπ‘₯ isn’t a fraction, we do treat it a little like one. We treat d𝑒 and dπ‘₯ as differentials. And we can say that d𝑒 equals dπ‘₯. Now, this may not seem instantly helpful, as if we replace dπ‘₯ with d𝑒 and π‘₯ plus five with 𝑒, we still have a part of our function that’s in terms of π‘₯.

However, if we look back to our substitution, we can rearrange this and say that π‘₯ must be equal to 𝑒 minus five. So, we can now replace each part of our integrand and dπ‘₯ with d𝑒. But what about our limits? Well, here we use our substitution to redefine these. Our lower limit is when π‘₯ is equal to negative one. And since 𝑒 was equal to π‘₯ plus five, 𝑒 is equal to negative one plus five, which is equal to four. Then, when π‘₯ is equal to four, that’s our upper limit, 𝑒 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 𝑒 minus five β€” remember, we said π‘₯ is equal to 𝑒 minus five β€” times the square root of 𝑒 with respect to 𝑒. And actually, let’s rewrite the square root of 𝑒 as 𝑒 to the power of one-half. And then, we can distribute the parentheses. When we multiply 𝑒 by 𝑒 to the power of one-half, we add their exponents. And we end up with 𝑒 to the power of three over two. So, our integrand is 𝑒 to the power of three over two minus five 𝑒 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 𝑒 to the power of three over two, we get 𝑒 to the power of five over two divided by five over two. And that’s the same as two-fifths times 𝑒 to the power of five over two. Similarly, when we integrate negative five 𝑒 to the power of one-half, we get negative five 𝑒 to the power of three over two divided by three over two. And that simplifies to negative 10 over three 𝑒 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.

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