# Video: Applying the Second Part of the Fundamental Theorem of Calculus

The graph shows the derivative of a function π. Given that π(3) = 2, what is π(5)?

03:06

### Video Transcript

The graph shows the derivative of a function π. Given that π of three is equal to two, what is π of four?

We need to be really careful here. Weβve not been given the graph of the function π but the graph of its derivative. Thatβs the graph of π prime. And so, we canβt simply read values straight off the graph. Instead, we need to find a way to link π prime with its values for π. Weβre going to use the second part of the fundamental theorem of calculus. And this says that if π is continuous on the closed interval π to π, then the definite integral between π and π of π of π₯ with respect to π₯ is equal to capital πΉ of π minus capital πΉ of π, where capital πΉ is the antiderivative of π. Thatβs a function such that capital πΉ prime is equal to π. Now, actually, we can reword this a little. Since π is the antiderivative of π prime, we can say that the definite integral between π and π of π prime of π₯ with respect to π₯ is equal to π of π minus π of π.

Now, before we apply this theorem, letβs just check the continuity of our function π. Well, weβre given a graph of its derivative and we know that any differentiable function must be continuous at every point in its domain. So, simply because weβve been given the derivative of π, we know π itself must be continuous. Now, weβre given π of three, and weβre looking to find π of four. So, weβre going to define π as being equal to three and π being equal to four. And this means then that the definite integral between three and four of π prime of π₯ with respect to π₯ must be equal to π of four minus π of three. Well, π of four is what weβre trying to find. But weβre told in the question that π of three is equal to two.

And what about the definite integral between three and four of π prime of π₯ with respect to π₯? Well, to work this out, we need to go to the graph. We know that the definite integral between two points of a function represents the area under a curve. And if that area lies below the π₯-axis, that will come out as a negative value. Weβre looking to find the area between the curve and the π₯-axis between the points π₯ equals three and π₯ equals four. Well, by dropping the vertical lines π₯ equals three and π₯ equals four onto our diagram, we see that we need to find the area of a trapezium. So, we can recall the area of a trapezium as being a half times π plus π times β, where π and π are the lengths of the parallel sides and β is the height between them.

Well, our two parallel sides are two units in length and four units in length. Thatβs essentially the absolute value of the value of the derivative of those points. The distance between them is one, and we know that the integral will give us a negative result since this trapezium lies below the π₯-axis. Well, two plus four is six, and six times one is six. So, we have negative one-half times six equals π of four minus two or negative three equals π of four minus two. Remember, weβre trying to find the value of π of four. So, weβre going to solve for π of four by adding two to both sides. π of four is therefore equal to negative three plus two, which is negative one. And so, by applying the second part of the fundamental theorem of calculus to the information weβve been given, we found π of four to be equal to negative one.