Question Video: Finding the Derivative of a Function Defined by an Integral Mathematics • Higher Education

Use the fundamental theorem of calculus to find the derivative of the function 𝑔(𝑠) = ∫_(1)^(𝑠) (3𝑑³ βˆ’ 4𝑑⁡)⁴ d𝑑.

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Video Transcript

Use the fundamental theorem of calculus to find the derivative of the function 𝑔 of 𝑠 is equal to the definite integral from one to 𝑠 of three 𝑑 cubed minus four 𝑑 to the fifth power all raised to the fourth power with respect to 𝑑.

We’re asked to find the derivative of the function 𝑔 of 𝑠. Because 𝑔 is a function in 𝑠, that’s the derivative of 𝑔 with respect to 𝑠. And we’re told to do this by using the fundamental theorem of calculus. So let’s start by recalling the fundamental theorem of calculus. This tells us if lowercase 𝑓 is a continuous function on a closed interval from π‘Ž to 𝑏 and capital 𝐹 of π‘₯ is the integral from π‘Ž to π‘₯ of lowercase 𝑓 of 𝑑 with respect to 𝑑, then capital 𝐹 prime of π‘₯ will be equal to lowercase 𝑓 of π‘₯ for all values of π‘₯ in the open interval from π‘Ž to 𝑏.

To apply the fundamental theorem of calculus to this problem, capital 𝐹 of π‘₯ is the function which we’re differentiating. In this case, that’s 𝑔 of 𝑠. So the first thing we should do is rewrite the fundamental theorem of calculus with π‘₯ replaced with 𝑠. And now we’re ready to start using the fundamental theorem of calculus. First, as we already explained, our function capital 𝐹 of π‘₯ will be 𝑔 of 𝑠. Next, in our integral, the lower limit of integration is one. This will be our constant π‘Ž. We can also see our upper limit of integration is 𝑠. Finally, our integrand will be lowercase 𝑓 of 𝑑.

Now, remember, to apply the fundamental theorem of calculus, we do need to check that our integrand lowercase 𝑓 is continuous on some closed interval containing one. In our case, the function lowercase 𝑓 of 𝑑 which is our integrand is equal to three 𝑑 cubed minus four 𝑑 to the fifth power all raised to the power of four. And we can see this is just a polynomial in 𝑑. This means it’s continuous for all real values of 𝑑. And in particular, this means it will be continuous on any closed interval containing one. Therefore, we’ve justified our use of the fundamental theorem of calculus.

Remember, capital 𝐹 of 𝑠 was our function 𝑔 of 𝑠 and lowercase 𝑓 of 𝑠 is our integrand. So the concluding statement of the fundamental theorem of calculus tells us 𝑔 prime of 𝑠 will be equal to three 𝑠 cubed minus four 𝑠 to the fifth power all raised to the power of four. And this is our final answer. Therefore, by applying the fundamental theorem of calculus to the function 𝑔 of 𝑠 is equal to the definite integral from one to 𝑠 of three 𝑑 cubed minus four 𝑑 to the fifth power all raised to the fourth power with respect to 𝑑, we were able to conclude that 𝑔 prime of 𝑠 will be equal to three 𝑠 cubed minus four 𝑠 to the fifth power all raised to the power of four.

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