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

Find ∫_(0)^(3) 2π‘₯/(9+2π‘₯Β²) dπ‘₯ to the nearest hundredth.

08:03

Video Transcript

Find the definite integral from zero to three of two π‘₯ over nine plus two π‘₯ squared with respect to π‘₯ to the nearest hundredth.

Now, there’re two ways that we could approach this question. The first way is to notice that the derivative with respect to π‘₯ of the denominator of this fraction, which is nine plus two π‘₯ squared, is four π‘₯. And that is a scalar multiple of the numerator of the fraction two π‘₯, which tells us that we could approach this question using the method of substitution. When we use the method of integration by substitution, we introduce a new variable. So we’re going to let 𝑒 equal nine plus two π‘₯ squared, the denominator of our fraction.

We’ve already seen that the derivative of nine plus two π‘₯ squared with respect to π‘₯ is four π‘₯. So we have that d𝑒 by dπ‘₯ equals four π‘₯. Now, d𝑒 by dπ‘₯ is absolutely not a fraction. But when we perform this method, we do treat it a little like one. So we can say that d𝑒 is equivalent to four π‘₯ dπ‘₯. Or if we wanted, we could divide both sides of this equation by two and say that a half d𝑒 is equivalent to two π‘₯dπ‘₯. Now, why is this helpful? Well, if we look back at the original integral, we see that we have two π‘₯ dπ‘₯. And we’ve just seen that this is equivalent to one-half d𝑒. We also said that nine plus two π‘₯ squared was equivalent to 𝑒.

So we know how to change everything inside of our integral to be in terms of 𝑒 instead of in terms of π‘₯. Replacing one of the nine plus two π‘₯ squared with one over 𝑒 then and two π‘₯dπ‘₯ with a half d𝑒, our integral becomes one over 𝑒 one-half d𝑒. But as this is a definite integral, we must remember to change the limits from limits in terms of π‘₯ to limits in terms of 𝑒. We use the relationship we’ve already defined between π‘₯ and 𝑒. 𝑒 is equal to nine plus two π‘₯ squared.

So for our lower limit, when π‘₯ is equal to zero, 𝑒 is equal to nine plus two times zero squared. That’s nine plus zero which is nine. And so, the lower limit for our integral in terms of 𝑒 is nine. For the upper limit, when π‘₯ is equal to three, 𝑒 is equal to nine plus two times three squared. That’s nine plus two times nine or nine plus 18 which is 27. And so, the upper limit for our integral in terms of 𝑒 is 27. We now have the definite integral from 9 to 27 of one over 𝑒 a half d𝑒. We can bring that constant factor of a half at the front of the integral if we wish. And we can write one over 𝑒 as 𝑒 to the power of negative one if it helps, giving one-half the integral from nine to 27 of 𝑒 to the power of negative one with respect to 𝑒.

Now, to perform this integration, we need to recall that when we integrate 𝑒 to the power of negative one with respect to 𝑒, we get a natural logarithm. The integral of 𝑒 to the power of negative one is the natural logarithm of the absolute value of 𝑒. So we have that the integral is equal to one-half the natural logarithm of the absolute value of 𝑒 evaluated between nine and 27. Substituting the limits gives a half the natural logarithm of the absolute value of 27 minus the natural logarithm of the absolute value of nine. But as 27 and nine are both positive values, their absolute values are just equal to themselves. So in fact, we can remove the absolute value signs.

Finally, we can use laws of logarithms to rewrite this as a half the natural logarithm of 27 over nine and 27 over nine is, of course, just three. So we have that this integral is equal to a half the natural logarithm of three. We were asked to give the answer to the nearest hundredth though. So evaluating using a calculator gives 0.5493. And to the nearest hundredth, that’s 0.55.

Notice that because we changed the limits in our integral to the limits in terms of 𝑒 rather than limits in terms of π‘₯, there was no need to reverse our substitution at the end.

Now, that’s one perfectly valid way of approaching this question: the method of integration by substitution. But in fact, there is a quicker way if we recognize that the integrand is in a particular form. We said right at the start of the question that the numerator of this fraction was a scalar multiple of the derivative of the denominator. So what we have is an integral of the form 𝑓 prime of π‘₯ over 𝑓 of π‘₯, with respect to π‘₯, subject to multiplication by a constant. We saw that this integral led to a natural logarithm.

And in fact, we can quote a general result, which is that if we’re integrating something of the form 𝑓 prime of π‘₯ over 𝑓 of π‘₯ with respect to π‘₯, then this is equal to the natural logarithm of the absolute value of 𝑓 of π‘₯. That’s the natural logarithm of the absolute value of the denominator of our fraction.

If the integral we were performing was indefinite, then we would have to add a constant of integration 𝑐. But as we’re working with a definite integral in this question, we can include the limits for the integral of π‘Ž and 𝑏. So if we can spot that our integral is of this type, subject to multiplication by a constant, then we can speed up the process. We already said that the derivative with respect to π‘₯ of nine plus two π‘₯ squared is four π‘₯. The numerator of our fraction is two π‘₯. So if we want to make it exactly the same as the derivative of the denominator, we’d need to multiply by two, which means we’d need to multiply the overall integral by a half so that we haven’t changed its value overall.

We can say then that the integral from zero to three of two π‘₯ over nine plus two π‘₯ squared with respect to π‘₯ is equivalent to a half the integral from zero to three of four π‘₯ over nine plus two π‘₯ squared with respect to π‘₯. And now, we have our function 𝑓 of π‘₯ in the denominator and its derivative 𝑓 prime of π‘₯ in the numerator. Applying our standard result then, we can say that this is equal to a half the natural logarithm of the absolute value of nine plus two π‘₯ squared, evaluated between zero and three.

Now, this looks quite different to the answer we got when we used the method of integration by substitution. But remember, we had changed our variable from π‘₯ to 𝑒. Substituting the limits for this integral gives a half the natural logarithm of the absolute value of nine plus two times three squared minus the natural logarithm of the absolute value of nine plus two times zero squared.

Simplifying gives a half the natural logarithm of the absolute value of 27 minus the natural logarithm of the absolute value of nine. And as both nine and 27 are positive, their absolute values are just equal to their own values. So we have a half the natural logarithm of 27 minus the natural logarithm of nine. And now, we see that this is identical to what we obtained doing our previous method. In the same way then, we can give an exact answer of a half the natural logarithm of three or a value 0.55 to the nearest hundredth.

Both of these methods are equally valid. But if you can spot integrands of the form 𝑓 prime of π‘₯ over 𝑓 of π‘₯ and recall the general result for integrating these or constant multiples of these. That is, that it’s equal to the natural logarithm of the absolute value of 𝑓 of π‘₯, the function in the denominator, then this will speed up the process.

Our answer to this problem is that the integral from zero to three of two π‘₯ over nine plus two π‘₯ squared with respect to π‘₯ to the nearest hundredth is 0.55.

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