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

Video Transcript

Find the derivative of the function 𝑔 of 𝑥 is equal to the integral between one minus two 𝑥 and one plus 𝑥 of five 𝑡 sin 𝑡 with respect to 𝑡.

For this question, we’ve been given a function defined by an integral, 𝑔 of 𝑥. And we need to find its derivative, 𝑔 prime of 𝑥. To do so we’ll be using the first part of the fundamental theorem of calculus, which tells us if we have an equation in this form, we can directly find its derivative using the following rules. Here, we have rewritten our equation. Setting 𝑓 of 𝑡 equal to five 𝑡 sin 𝑡. Now, the first thing we will notice when looking at our integral is that not only other top and bottom limits functions of 𝑥 instead of 𝑥 itself. But that of them are a constant, which we need to use our theorem.

At this point, we can recall that an integral can be split in the following way. To conceptualize this, it can be helpful to interpret a definite integral is the area under a curve as shown. Using this method we can artificially introduce a constant which we’ll call 𝑎 into the integrals which now form our sum. Next, we noticed that our first integral has this constant as its upper limit instead of its lower limit. And we need to switch things round to use the fundamental theorem of calculus. Luckily, switching the bounds of an integral and multiplying by minus one evaluates to the same result. And so we can do this.

Okay, we’re a bit closer to the form we need now. However, remember, the upper limit of both of these integrals instead of being 𝑥 is a function of 𝑥. To move forward, we can use the following modified form of the fundamental theorem of calculus. If lowercase 𝑓 is a continuous function over the closed interval between 𝑎 and 𝑏, and uppercase 𝐹 of 𝑥 is defined by the integral between 𝑎 and some function of 𝑥, which we’ll call 𝑢 of 𝑥 of 𝑓of 𝑡 with respect to 𝑡. Then capital 𝐹 prime of 𝑥 is equal to lower case 𝑓 of 𝑢 of 𝑥 multiplied by d by d𝑥 of 𝑢 of 𝑥, where 𝑢 of 𝑥 is in the open interval between 𝑎 and 𝑏. Okay, so looking back at 𝑔 of 𝑥, it is now expressed is the sum of two separate integrals.

Given the rules of differentiation, we confined 𝑔 prime of 𝑥 simply by differentiating each of these terms individually. And therefore, our modified theorem can also be applied individually to both of the terms. For the first term our upper limit is the function one minus two 𝑥. Our modified theorem tells us that this is, therefore, equal to negative 𝑓 of one minus two 𝑥 times d by d𝑥 of one minus two 𝑥. Our second term takes the same form, but using the upper limit of one plus 𝑥. We now remember that 𝑓 of 𝑡 is equal to five 𝑡 sin 𝑡. For 𝑓 of one minus two 𝑥, we’ll replace all of the 𝑡 terms in this equation with one minus two 𝑥. We must then multiply this by the derivative of one minus two 𝑥.

For our second term, we follow exactly the same pattern, first substituting in to lowercase 𝑓 and then finding the derivative of one plus 𝑥 and multiplying. After a bit of simplification, we’re left with the following result. We have now completed our question and we have found the derivative 𝑔 prime of 𝑥. This example illustrates that we can use the modified version of the fundamental theorem of calculus even when the function that we have is defined by an integral involving two different functions of 𝑥 in the upper and lower limits.