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Is the integral between zero and five of 𝑥 divided by 𝑥 squared minus 16 with respect to 𝑥 convergent?
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If so, what does it converge to?
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For this question, we’ve been given an improper integral with a discontinuous integrand.
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To move forward, we’ll need to see where these discontinuities occur and to see how they relate to our limits of integration.
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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.
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Solving this equation, we find that our denominator is equal to zero when 𝑥 is equal to positive or negative four.
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Okay, let’s see how these values relate to our limits of integration, zero and five.
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Our first value, negative four, is not equal to either of the limits of integration, nor does it lie between them.
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The value of four, however, does lie between our limits of integration.
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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.
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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.
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So we’ve ignored the discontinuity at 𝑥 equals negative four since it does not interact with our limits of integration.
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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.
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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.
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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.
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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.
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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.
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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.
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We’ll therefore need to check our two smaller integrals for convergence.
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However, before we do this, the technique that we’ll be using to solve these two integrals will be a 𝑢-substitution.
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And so we might as well get this out of the way first.
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The substitution that we’ll be using is that 𝑢 is equal to 𝑥 squared minus 16.
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Differentiating with respect to 𝑥, we get that d𝑢 by d𝑥 is equal to two 𝑥.
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And an equivalent statement for this is that d𝑢 is equal to two 𝑥 d𝑥.
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It turns out that a more useful equation for us is that half d𝑢 is equal to 𝑥 d𝑥.
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Okay.
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Since we’re working with definite integrals, we should also pay attention to our limits of integration.
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When 𝑥 is equal to zero, 𝑢 is equal to negative 16.
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When 𝑥 is four, 𝑢 is zero.
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And when 𝑥 is five, 𝑢 is nine.
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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.
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After performing these substitutions, we also might as well take this factor of a half outside of the integrals.
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Okay.
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At this stage, it’s relevant to give a quick but very important side note.
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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.
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We do this because performing substitutions first can sometimes cause problems by removing the discontinuity.
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Although this sounds great, in practice, it can lead to some problems.
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To move forward, we’re gonna need to check our integrals for convergence.
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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.
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We’ll begin with the integral with a discontinuity at the upper limit.
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The definition of an improper integral gives us the following tool to deal with this case.
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Applying this to our first term gives us the following result.
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To move forward, we use the fact that the antiderivative of one over 𝑢 is the natural logarithm of the absolute value of 𝑢.
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We’ll get rid of these definitions to clear some room.
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We input our limits of integration.
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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.
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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.
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This is another way of saying that the limit does not exist.
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Since the limit that defines the first of our smaller integrals does not exist, we say it is divergent.
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And in fact, this divergence cascades all the way back to our original integral.
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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.
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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.
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With no further work, we can say that the integral given by the question is not convergent, but rather divergent.