Question Video: Finding the Derivative of a Function Defined by an Integral Where Its Limits Contain a Variable Mathematics • Higher Education

Video Transcript

Find the derivative of the function capital 𝐹 of 𝑥 is equal to the definite integral from five 𝑥 to two 𝑥 squared of two 𝑒 to the power of 𝑡 squared with respect to 𝑡.

We’re given a function capital 𝐹 of 𝑥 which is a function in 𝑥. And we can see that this function is the definite integral where both of our limits of integration are functions in 𝑥. We need to find the derivative of this function capital 𝐹 of 𝑥 that would be with respect to 𝑥. And since this is a function defined by an integral where the limits of integration involve 𝑥, we’ll do this by using the fundamental theorem of calculus.

So let’s recall the fundamental theorem of calculus. In fact, we’ll only recall a part of this. This tells us if lowercase 𝑓 is a continuous function on the closed interval from 𝑎 to 𝑏 and capital 𝐹 of 𝑥 is equal to the definite integral from 𝑎 to 𝑥 of lowercase 𝑓 of 𝑡 with respect to 𝑡, then capital 𝐹 prime of 𝑥 will be equal to lowercase 𝑓 of 𝑥 for all values of 𝑥 in the open interval from 𝑎 to 𝑏. So this gives us a method of differentiating a function defined by an integral where the upper limit of integration is 𝑥. And this means we can immediately see a problem.

In the function given to us in the question, the upper limit of integration is not just 𝑥. In fact, it’s a function of 𝑥. We have two 𝑥 squared. And the lower limit of integration is supposed to be a constant. However, in our case, it’s also a function in 𝑥. So to use the fundamental theorem of calculus, we’re going to first need to rewrite our function capital 𝐹 of 𝑥 in this form. And the question is, how are we going to rewrite our function capital 𝐹 of 𝑥?

To do this, we’re going to need to recall one of our rules for definite integrals. We need to recall that we can split our integral into two at the constant value of 𝑐. But we need to be careful here because this will only be true if our integrand is continuous over both domains of integration. So before we do this, let’s check where our integrand is continuous. In this case, our integrand is two times 𝑒 to the power of 𝑡 squared. This is a composition between a polynomial and an exponential function. And we know that polynomials are continuous for all real values, and the exponential function is also continuous for all real values.

This means that our integrand is the composition of continuous functions and must be continuous across its entire domain. And we can see the domain of our integrand is all real values. And therefore, it’s continuous for all real values and must be integral across any interval. So in this case, we can pick any value for our constant 𝑐 because our integrand is integrable on any interval. We’ll use 𝑐 is equal to zero. However, it won’t change the answer in our question. We’ve now split our definite integral into two integrals. However, neither of these are in a form which we can directly use the fundamental theorem of calculus. So we need to modify these even more.

Let’s take a look at our second integral. We can see the upper limit of integration is a constant. However, we want the lower limit of integration to be a constant, so we’re going to need to use another one of our rules for definite integrals. We need to recall that we can switch our upper and lower limits of integration if we multiply our integral by negative one. So by using this, we now subtract the integral from zero to five 𝑥 of two 𝑒 to the power of 𝑡 squared with respect to 𝑡. Now, both of these integrals are almost in a form we can directly use the fundamental theorem of calculus. The only problem is in both of these integrals, the upper limit of integration is not 𝑥; it’s a function in 𝑥.

There are two ways we could approach this problem. Let’s focus on our first integral. We could use the substitution 𝑣 of 𝑥 is equal to two 𝑥 squared. Then our definite integral would now be a function in 𝑣. And we want to differentiate this with respect to 𝑥. So we want to differentiate a function in 𝑣 where 𝑣, in turn, is a function in 𝑥. We could do this by using the chain rule. And we could do exactly the same for our second integral. However, we can actually do this in the general case to our fundamental theorem of calculus.

If 𝑣 of 𝑥 is a differentiable function, lowercase 𝑓 is continuous on the closed interval from 𝑎 to 𝑏, and capital 𝐹 of 𝑥 is equal to the definite integral from 𝑎 to 𝑣 of 𝑥 of lowercase 𝑓 of 𝑡 with respect to 𝑡, then by differentiating both sides of this equation with respect to 𝑥, using the chain rule and using our standard statement of the fundamental theorem of calculus, we get that capital 𝐹 prime of 𝑥 will be equal to lowercase 𝑓 evaluated at 𝑣 of 𝑥 times 𝑣 prime of 𝑥 for all values of 𝑣 of 𝑥 in the open interval from 𝑎 to 𝑏.

And we can now use this to differentiate each of our integrals separately. We’ll set 𝑣 one of 𝑥 to be two 𝑥 squared and 𝑣 two of 𝑥 to be five 𝑥. It’s worth remembering that these need to be differentiable functions because we need to use the chain rule. In this case, both of these are polynomial. So we know they’re differentiable. There’s one more thing we need to do before we use our modified version of the fundamental theorem of calculus. In both of our integrals, the lower limit of integration is zero. So we set lowercase 𝑎 equal to zero, and we need to check where our integrand lowercase 𝑓 is continuous.

We’ve already done this, though. We already explained that our integrand is continuous for all real values. So we already know that this is continuous on any closed interval. We’re now ready to differentiate both sides of this equation with respect to 𝑥 by using our modified version of the fundamental theorem of calculus. We just differentiate each of our integrals individually. We get capital 𝐹 prime of 𝑥 is equal to lowercase 𝑓 evaluated at 𝑣 one of 𝑥 multiplied by 𝑣 one prime of 𝑥 minus lowercase 𝑓 evaluated at 𝑣 two of 𝑥 multiplied by 𝑣 two prime of 𝑥.

And now, we can start evaluating this expression. Let’s start with 𝑓 evaluated at 𝑣 one of 𝑥. We just substitute 𝑡 is equal to two 𝑥 squared into our integrand. This gives us two 𝑒 to the power of two 𝑥 squared all squared. Next, we want to multiply this by 𝑣 one prime of 𝑥. That’s the derivative of two 𝑥 squared with respect to 𝑥. We can do this by using the power rule for differentiation; we get four 𝑥. And we can do the same for our second term. We substitute five 𝑥 into our integrand, giving us two times 𝑒 to the power of five 𝑥 all squared. And we want to multiply this by 𝑣 two prime of 𝑥. 𝑣 two of 𝑥 is five 𝑥. And the derivative of five 𝑥 with respect to 𝑥 is five.

And finally, we’ll simplify this expression. We’re going to evaluate our exponents and simplify our coefficients. This gives us capital 𝐹 prime of 𝑥 is equal to eight 𝑥𝑒 to the power of four 𝑥 to the fourth power minus 10𝑒 to the power of 25𝑥 squared. And this is our final answer.

Therefore, by using what we know about definite integrals, the chain rule, and the fundamental theorem of calculus, we were able to show if capital 𝐹 of 𝑥 is the definite integral from five 𝑥 to two 𝑥 squared of two 𝑒 to the power of 𝑡 squared with respect to 𝑡, then capital 𝐹 prime of 𝑥 is equal to eight 𝑥𝑒 to the power of four 𝑥 to the fourth power minus 10𝑒 to the power of 25𝑥 squared.