Question Video: Deciding If a Given Definite Integral Is Convergent Mathematics • Higher Education

Is the integral ∫_(0) ^(5) 𝑥/(𝑥² − 16) d𝑥 convergent? If so, what does it converge to?


Video Transcript

Is the integral between zero and five of 𝑥 divided by 𝑥 squared minus 16 with respect to 𝑥 convergent? If so, what does it converge to?

For this question, we’ve been given an improper integral with a discontinuous integrand. To move forward, we’ll need to see where these discontinuities occur and to see how they relate to our limits of integration. Since our integrand has been given in the form of a quotient, we can find where its discontinuities occur by finding where its denominator, 𝑥 squared minus 16, is equal to zero. Solving this equation, we find that our denominator is equal to zero when 𝑥 is equal to positive or negative four. Okay, let’s see how these values relate to our limits of integration, zero and five. Our first value, negative four, is not equal to either of the limits of integration, nor does it lie between them. The value of four, however, does lie between our limits of integration. For due diligence, we can check our integrand, which we’ll call 𝑓 of 𝑥, for the type of discontinuity that we expect to see when 𝑥 equals four. Trying to evaluate 𝑓 of four, we’re left with four over zero, which implies that we’re looking at an infinite discontinuity, as opposed to if it’d been zero over zero, which would imply a removable discontinuity.

So we’ve ignored the discontinuity at 𝑥 equals negative four since it does not interact with our limits of integration. We’ve also confirmed that the question has given us an improper integral with a discontinuity occurring between the limits of integration, which is when 𝑥 equals four. The definition of an improper integral gives us the following tool, which allows us to split an integral at the discontinuity, 𝑐, into the sum of two smaller integrals. Let us apply this to our question, where we have our integrand 𝑓 of 𝑥, our lower limit 𝑎, our upper limit 𝑏, and our discontinuity 𝑐. Using our definition, we can say that our original integral is equal to the sum of two smaller integrals, which are adjacent to each other at the discontinuity when 𝑥 is equal to four. We can see this in the fact that the discontinuity occurs at the upper limit of our first integral and the lower limit of our second integral.

At this point, we should note that this statement is only valid and our original integral is only convergent if both of the two smaller integrals themselves are convergent. We’ll therefore need to check our two smaller integrals for convergence. However, before we do this, the technique that we’ll be using to solve these two integrals will be a 𝑢-substitution. And so we might as well get this out of the way first. The substitution that we’ll be using is that 𝑢 is equal to 𝑥 squared minus 16. Differentiating with respect to 𝑥, we get that d𝑢 by d𝑥 is equal to two 𝑥. And an equivalent statement for this is that d𝑢 is equal to two 𝑥 d𝑥. It turns out that a more useful equation for us is that half d𝑢 is equal to 𝑥 d𝑥.

Okay. Since we’re working with definite integrals, we should also pay attention to our limits of integration. When 𝑥 is equal to zero, 𝑢 is equal to negative 16. When 𝑥 is four, 𝑢 is zero. And when 𝑥 is five, 𝑢 is nine. So let us now perform our substitutions, replacing 𝑥 squared minus 16 with 𝑢, 𝑥 d𝑥 with a half d𝑢, and our limits of integration as we’ve just found. After performing these substitutions, we also might as well take this factor of a half outside of the integrals.

Okay. At this stage, it’s relevant to give a quick but very important side note. When working with an improper integral which contains a discontinuity between the limits of integration, we should always split the integral at the discontinuity before performing any substitutions. We do this because performing substitutions first can sometimes cause problems by removing the discontinuity. Although this sounds great, in practice, it can lead to some problems. To move forward, we’re gonna need to check our integrals for convergence. We note that since our integrals have the discontinuity at the upper and lower limits, respectively, before the substitution, the same is true after the substitution. We’ll begin with the integral with a discontinuity at the upper limit.

The definition of an improper integral gives us the following tool to deal with this case. Applying this to our first term gives us the following result. To move forward, we use the fact that the antiderivative of one over 𝑢 is the natural logarithm of the absolute value of 𝑢. We’ll get rid of these definitions to clear some room. We input our limits of integration. We can then take a shortcut by noticing that if we try to evaluate this first term, we would be left with a negative infinity. Since our other term, which is the natural logarithm of the absolute value of negative 16, is finite, we’re forced to conclude that our original limit will evaluate to a negative infinity. This is another way of saying that the limit does not exist. Since the limit that defines the first of our smaller integrals does not exist, we say it is divergent. And in fact, this divergence cascades all the way back to our original integral.

Recall that our definition told us that the original relationship we used is only valid if both of the two smaller integrals, which we’re summing, are convergent. Since the first one we checked turned out to be divergent, we don’t actually need to check the second one to conclude that our original integral is also divergent itself. With no further work, we can say that the integral given by the question is not convergent, but rather divergent.

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