Question Video: Finding the Derivative of a Function Defined by an Integral Where the Lower Limit Is a Variable Mathematics • Higher Education

Video Transcript

Use the fundamental theorem of calculus to find the derivative of the function 𝑅 of 𝑦 is equal to the integral between 𝑦 and five of three 𝑡 squared sin two 𝑡 with respect to 𝑡.

For this question, we first recognize that we’ve been given a function defined by an integral capital 𝑅 of 𝑦. And we’ve been asked to find the derivative of this, capital 𝑅 prime of 𝑦. A tool that we can use to solve this problem, as the question tells us, is the fundamental theorem of calculus. The first part of this tells us that if lowercase 𝑓 is a continuous function on the closed interval between 𝑎 and 𝑏 and the function capital 𝐹 of 𝑥 is defined by the integral between 𝑎 and 𝑥 of lowercase 𝑓 of 𝑡 with respect to 𝑡. Then 𝐹 prime of 𝑥 is equal to 𝑓 of 𝑥 for all values of 𝑥 on the open interval between 𝑎 and 𝑏. Now, the equation given in our question does not have a capital 𝐹 of 𝑥, but instead we have a capital 𝑅 of 𝑦.

Before we start to think about 𝑅 prime of 𝑦, let’s dial things back to 𝑅 of 𝑦 to see if we have the correct form of equation to apply the fundamental theorem of calculus. We do indeed have a function defined by an integral and a continuous function as our integrand. However, the variable that we’re operating on, 𝑦, appears as the lower limit of integration, and we have a constant as the upper limit. This is the other way round to the fundamental theorem of calculus, where the variable is the upper limit and the constant is the lower limit. This means the positions of our limits are reversed. Luckily, one of the properties of integrals is that switching the positions of the limits and multiplying by negative one is equal to the original Integral. We can do this to the integral, which defines 𝑅 of 𝑦. And of course, it doesn’t matter whether this factor of negative one is inside or outside of our integral.

Now that our equation is in the correct form with a variable appearing as the upper limit and the constant appearing as the lower limit, we can use the fundamental theorem of calculus. This allows us to say that capital 𝑅 prime of 𝑦 is equal to 𝑓 of 𝑦. And remember the variable is 𝑦 not 𝑥. Looking back at our integral, we have to find lowercase 𝑓 of 𝑡 to be negative three 𝑡 squared sin two 𝑡. This means that for lowercase 𝑓 of 𝑦 we’ll replace all of our 𝑡s for 𝑦s. We, therefore, find that capital 𝑅 prime of 𝑦 is equal to negative three 𝑦 squared sin two 𝑦. And we have, therefore, answered our question. We have found the derivative capital 𝑅 prime of 𝑦 using the fundamental theorem of calculus. And this allowed us to avoid evaluating the definite integral given in the question which may have led to messy or lengthy calculations.