Question Video: Evaluating the Improper Integral of a Discontinuous Function | Nagwa Question Video: Evaluating the Improper Integral of a Discontinuous Function | Nagwa

# Question Video: Evaluating the Improper Integral of a Discontinuous Function Mathematics • Higher Education

The integral ∫_(0) ^(1) 1/√(1 − 𝑥²) d𝑥 is convergent. What does it converge to?

03:59

### Video Transcript

The integral from zero to one of one divided by the square root of one minus 𝑥 squared with respect to 𝑥 is convergent. What does it converge to?

We’re given a definite integral, and we’re told that this integral is convergent. We need to determine what value does this integral converge to. And to do this, we need to notice something about our integral. In fact, this integral is an improper integral. First, our integrand is the composition of continuous functions. It’s a power function composed with a quadratic. And the composition of continuous functions is continuous across its entire domain. And we can find the domain of our integrand. Remember, we’re not allowed to divide by zero, and we’re not allowed to take the square root of a negative number. So our integrand will be defined and therefore continuous everywhere where one minus 𝑥 squared is greater than zero.

And we can then solve this inequality for 𝑥. First, we’ll add 𝑥 squared to both sides; we get one must be greater than 𝑥 squared. And then we can solve this by using a graphical approach or by using what we know about the square root. We get that 𝑥 must be between negative one and one. And this is where we can see our problem. We’re using the integral from zero to one. This means we need our integrand to be continuous on the closed interval from zero to one. However, we’ve just shown that our integrand is not continuous when 𝑥 is equal to one. This is because we’ll be dividing by zero. So we’re going to need to use what we know about improper integrals.

We need to recall the following rule for integrating improper integrals. If 𝑓 is a continuous function on a closed interval from 𝑎 to 𝑏 except at the value of 𝑏 where 𝑓 of 𝑥 is discontinuous, then we know the integral from 𝑎 to 𝑏 of 𝑓 of 𝑥 with respect to 𝑥 is equal to the limit as 𝑡 approaches 𝑏 from the left of the integral from 𝑎 to 𝑡 of 𝑓 of 𝑥 with respect to 𝑥 as long as this limit exists and is finite. In other words, this gives us a method of turning an improper integral into the limit of an integral which we know will be proper. And this is exactly what we have in the integral given to us in the question. We want to set our value of 𝑎 equal to zero and 𝑏 equal to one.

In fact, we can update this definition for improper integrals with 𝑎 set to be zero and 𝑏 set to be one. So if we set 𝑓 to be the integrand given to us in the question, we’ve shown that 𝑓 is continuous if 𝑥 is between negative one and one. In particular, this means it’s continuous for values of 𝑥 greater than or equal to zero and less than one. And we’ve also shown that our function is not defined when 𝑥 is equal to one. So in particular, this means it’s discontinuous when 𝑥 is equal to one. So our integral is improper. And we should try to evaluate it by using this limit.

So we now have the integral from zero to one of one divided by the square root of one minus 𝑥 squared with respect to 𝑥 is equal to the limit as 𝑡 approaches one from the left of the integral from zero to 𝑡 of one divided by the square root of one minus 𝑥 squared with respect to 𝑥 as long as this limit exists. And one thing worth pointing out is we’re now integrating from zero to 𝑡. And remember, 𝑡 is approaching one from the left; it’s never equal to one. So our integrand is now continuous on the interval of integration. It’s now a proper integral, so we can use any of our tools to evaluate this integral.

To evaluate this integral, we need to notice something about our integrand. We need to recall the derivative of the inverse sin of 𝑥 with respect to 𝑥 is equal to one divided by the square root of one minus 𝑥 squared. Therefore, the inverse sin of 𝑥 is an antiderivative of our integrand. So we can use this to evaluate our integral. This gives us the limit as 𝑡 approaches one from the left of the inverse sin of 𝑥 evaluated at the limits of integration zero and 𝑡. And now all we need to do is evaluate this at the limits of integration. Doing this, we get the limit as 𝑡 approaches one from the left of the inverse sin of 𝑡 minus the inverse sin of zero.

And we can evaluate this expression. First, the inverse sin of zero is just equal to zero. Then, all we need to remember is the inverse sin of 𝑡 is continuous when 𝑡 is equal to one, so we can just evaluate this by using direct substitution. So this limit evaluates to give us the inverse sin of one, which we can calculate is equal to 𝜋 by two, which is our final answer.

Therefore, we were able to show the integral from zero to one of one divided by the square root of one minus 𝑥 squared converges to 𝜋 by two.

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