### Video Transcript

The table shows some values of a differentiable function π and its derivative π prime. Find the definite integral between one and four of π prime of π₯ with respect to π₯.

To answer this question, weβre going to recall the second part of the fundamental theorem of calculus. This says that if π is a real-valued function on the closed interval π to π such that capital πΉ is an antiderivative of π on this same interval. Then if π is Reimann integrable on that closed interval, then the definite integral between π and π of π of π₯ with respect to π₯ is equal to capital πΉ of π minus capital πΉ of π.

We can see that π of π₯ and π prime of π₯ are real-valued functions on the closed interval one to four. Now, it also follows that if π prime of π₯ is the derivative of π of π₯, then π of π₯ must itself be the antiderivative of π prime of π₯. We can use this to rewrite our theorem such that the definite integral between π and π of π prime of π₯ with respect to π₯ must be equal to π of π minus π of π. In our case, this means that the definite integral between one and four of π prime of π₯ with respect to π₯ must be π of four minus π of one. We can see from our table that π of four is nine and π of one is two.

So the definite integral between one and four of π prime of π₯ with respect to π₯ is nine minus two, which is equal to seven.