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.
