# 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 π(π ) = β«_(1)^(π ) (3π‘Β³ β 4π‘β΅)β΄ dπ‘.

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### Video Transcript

Use the fundamental theorem of calculus to find the derivative of the function π of π  is equal to the definite integral from one to π  of three π‘ cubed minus four π‘ to the fifth power all raised to the fourth power with respect to π‘.

Weβre asked to find the derivative of the function π of π . Because π is a function in π , thatβs the derivative of π with respect to π . And weβre told to do this by using the fundamental theorem of calculus. So letβs start by recalling the fundamental theorem of calculus. This tells us if lowercase π is a continuous function on a closed interval from π to π and capital πΉ of π₯ is the integral from π to π₯ of lowercase π of π‘ with respect to π‘, then capital πΉ prime of π₯ will be equal to lowercase π of π₯ for all values of π₯ in the open interval from π to π.

To apply the fundamental theorem of calculus to this problem, capital πΉ of π₯ is the function which weβre differentiating. In this case, thatβs π of π . So the first thing we should do is rewrite the fundamental theorem of calculus with π₯ replaced with π . And now weβre ready to start using the fundamental theorem of calculus. First, as we already explained, our function capital πΉ of π₯ will be π of π . Next, in our integral, the lower limit of integration is one. This will be our constant π. We can also see our upper limit of integration is π . Finally, our integrand will be lowercase π of π‘.

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

Remember, capital πΉ of π  was our function π of π  and lowercase π of π  is our integrand. So the concluding statement of the fundamental theorem of calculus tells us π prime of π  will be equal to three π  cubed minus four π  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 π of π  is equal to the definite integral from one to π  of three π‘ cubed minus four π‘ to the fifth power all raised to the fourth power with respect to π‘, we were able to conclude that π prime of π  will be equal to three π  cubed minus four π  to the fifth power all raised to the power of four.