Video Transcript
The integral from zero to nine of one divided by the cube root of 𝑥 minus one with respect to 𝑥 is convergent. What does it converge to?
We’re given a definite integral which we are told is convergent. We need to find what this integral converges to.
First, let’s take a look at our integral. We can see that our integrand is one divided by the cube root of 𝑥 minus one. We know we can’t divide by zero. So, our integrand won’t be defined when 𝑥 is equal to one. In fact, we can say something more. Our integrand is the composition of continuous functions; it’s a linear function which we take the cube root of and then we take the reciprocal. And we know the composition of continuous functions is continuous on its domain.
Since we want the definite integral from zero to nine, we want to know where our integral is continuous on the closed interval from zero to nine. And we know the only place our integrand is not defined is when 𝑥 is equal to one. So, it’s continuous on the entire closed interval except when 𝑥 is equal to one. So, we need to work out how to deal with this improper integral where the discontinuity is in the middle of our interval.
Well, if our integral is convergent, then we can use our properties of definite integrals to split this integral into two different integrals. We’ll split it so that the upper limit of our first integral is one and the lower limit of our second integral is one.
Now, in our first integral, our integrand is continuous everywhere except at the upper limit of one. And in our second integral, the integrand is continuous everywhere except the lower limit of one. This means we can try and evaluate each of these integrals separately by using our rules for improper integrals.
Let’s talk about our first integral. The integrand 𝑓 is continuous everywhere except at the upper limit of our integral 𝑏. In fact, it has a discontinuity at 𝑏. This means by using our rules of improper integrals, we can evaluate the integral from 𝑎 to 𝑏 of 𝑓 of 𝑥 with respect to 𝑥 as the limit as 𝑡 approaches 𝑏 from the left of the integral from 𝑎 to 𝑡 of 𝑓 of 𝑥 with respect to 𝑥. And that’s, of course, if this limit exists. So, applying this to our first integral, we can rewrite it as the limit as 𝑡 approaches one from the left of the integral from zero to 𝑡 of one divided by the cube root of 𝑥 minus one with respect to 𝑥 if this limit exists.
We now need to do something similar with our second Integral. In our second integral, our integral 𝑓 is continuous everywhere except when 𝑥 is equal to the lower limit of our integral 𝑎. In fact, our integrand is discontinuous at 𝑥 is equal to 𝑎. This means we can evaluate our integral from 𝑎 to 𝑏 of 𝑓 of 𝑥 with respect to 𝑥 as the limit as 𝑡 approaches 𝑎 from the right of the integral from 𝑡 to 𝑏 of 𝑓 of 𝑥 with respect to 𝑥 as long as this limit exists. Applying this to our second integral, we get the limit as 𝑡 approaches one from the right of the integral from 𝑡 to nine of one divided by the cube root of 𝑥 minus one with respect to 𝑥 as long as this limit exists.
So, let’s clear some space and work on evaluating these limits. We now have to evaluate two definite integrals. We’ll do both of these by using a 𝑢 substitution. We’ll use 𝑢 is equal to 𝑥 minus one. If 𝑢 is equal to 𝑥 minus one, differentiating both sides with respect to 𝑥, we get d𝑢 by d𝑥 is equal to one. And we know that d𝑢 by d𝑥 is not a fraction. However, when we’re using integration by substitution, we can treat it a little bit like a fraction. This gives us the equivalent statement d𝑢 is equal to d𝑥.
The last thing we need to do is since we’re calculating a definite integral, we need to change the limits of our integral. We’ll start with the lower limit of our first integral. That’s when 𝑥 is equal to zero. We substitute this into our expression for 𝑢. We get 𝑢 is equal to zero minus one, which is equal to negative one. Let’s now do the upper limit of our first integral. That’s when 𝑥 is equal to 𝑡. We substitute this into our expression for 𝑢, and we get 𝑢 is equal to 𝑡 minus one.
This means we’ve found all the information we need to rewrite our first integral by using our 𝑢 substitution. We get the limit as 𝑡 approaches one from the left of the integral from negative one to 𝑡 minus one of one divided by the cube root of 𝑢 with respect to 𝑢. Our second integral is exactly the same except we have different limits for our integral.
We’ve actually already found the lower limit of this integral. When 𝑥 is equal to 𝑡, 𝑢 is equal to 𝑡 minus one. So, we just need to find the value of 𝑢 when 𝑥 is equal to nine. We get 𝑢 is equal to nine minus one, which we can calculate is equal to eight. So, by using our 𝑢 substitution on our second term, we get the limit as 𝑡 approaches one from the right of the integral from 𝑡 minus one to eight of one divided by the cube root of 𝑢 with respect to 𝑢.
This is almost in a form which we can integrate. We’ll just use our laws of exponents to rewrite one divided by the cube root of 𝑢 as 𝑢 to the power of negative one over three. We can now evaluate both of these integrals by using the power rule for integration. We want to add one to our exponent of 𝑢 and then divide by this new exponent of 𝑢.
Integrating 𝑢 to the power of negative one-third by using our power rule for integration, we add one to our exponent to give us two-thirds then we divide by this exponent of two-thirds. Dividing by two-thirds is the same as multiplying by three over two. This gives us the following expression. All we need to do now is evaluate both of these at the limits of their integrals. Evaluating these expressions at the limits of their integrals. We get the limit as 𝑡 approaches one from the left of three times 𝑡 minus one to the power of two over three divided by two minus three times negative one to the power of two over three divided by two plus the limit as 𝑡 approaches one from the right of three times eight to the power of two over three divided by two minus three times 𝑡 minus one to the power of two over three divided by two.
So, let’s clear some space so we can evaluate these limits. The first thing we need to notice is three times 𝑡 minus one to the power of two over three divided by two is a continuous function. And it’s a continuous function because it’s the composition of continuous functions. This means we can evaluate its limit by using direct substitution. In fact, the only other terms in our limit are constants.
Evaluating this limit by using direct substitution, we get three times one minus one to the power of two over three divided by two minus three times negative one to the power of two over three divided by two plus three times eight to the power of two over three divided by two minus three times one minus one to the power of two over three divided by two. And now, we just need to evaluate this expression.
Our first term and our last term contain a factor of one minus one, which is equal to zero. So, our first and last term evaluate to give us zero. Next, raising a number to the power of two over three is the same as taking the cube root and then squaring it. The cube root of negative one is negative one. And then, negative one squared is one. This means negative three times negative one to the power of two over three divided by two is just equal to negative three over two.
By using a similar logic, the cube root of eight is two. And then, two squared is four. So, eight to the power of two over three is equal to four. So, our third term evaluates to give us six. And finally, we can just evaluate this expression. We get nine divided by two. Therefore, because our limits converged and they converged to nine over two, we’ve shown the integral from zero to nine of one divided by the cube root of 𝑥 minus one is convergent and it converges to nine divided by two.
