## Introduction

The Second Fundamental Theorem of Calculus establishes a relationship between a function and its anti-derivative. Specifically, for a function f that is continuous over an interval I containing the x-value a, the theorem allows us to create a new function, F(x), by integrating f from a to x. When we do this, F(x) is the anti-derivative of f(x), and f(x) is the derivative of F(x). By this point, you probably know how to evaluate both derivatives and integrals, and you understand the relationship between the two. So while this relationship might feel like no big deal, the Second Fundamental Theorem is a powerful tool for building anti-derivatives when there seems to be no simple way to do so. Let’s get to the specifics.

## The Definition of the Second Fundamental Theorem of Calculus

Assume that f(x) is a continuous function on the interval I, which includes the x-value a. The Second Fundamental Theorem of Calculus defines a new function, F(x):

F(x)=\int _{ a }^{ x }{ f(t)dt }

where F(x) is an anti-derivative of f(x) for all x in I. That is, F'(x)=f(x).

Further, F(x) is the accumulation of the area under the curve f from a to x.

## Sample Questions

#### Example 1

Find F'(x), given F(x)=\int _{ -3 }^{ x }{ { t }^{ 2 }+2t-1dt }.

This is a very straightforward application of the Second Fundamental Theorem of Calculus. Since the lower limit of integration is a constant, -3, and the upper limit is x, we can simply take the expression { t }^{ 2 }+2t-1 given in the problem, and replace t with x in our solution. The solution to the problem is, therefore,

F'(x)={ x }^{ 2 }+2x-1 .

#### Example 2

Find F'(x), given F(x)=\int _{ -1 }^{ x^{ 2 } }{ -2t+3dt }.

Here, we will apply the Second Fundamental Theorem of Calculus. The lower limit of integration is a constant (-1), but unlike the prior example, the upper limit is not x, but rather { x }^{ 2 }. Thus, the integral as written does not match the expression for the Second Fundamental Theorem of Calculus upon first glance. We can work around this by making a substitution.

We let the upper limit of integration equal u. That is, we let u={ x }^{ 2 }. That gives us

F(x)=\int _{ -1 }^{ u }{ -2t+3dt }.

Next, we invoke the following equality from the chain rule: \dfrac { dF }{ dx } =\dfrac { dF }{ du } \cdot \dfrac { du }{ dx }. To use this equality, let’s focus on the right hand side.

\dfrac { dF }{ du } is the derivative of the given function, F, with respect to the new variable, u, that we have just introduced.

Meanwhile, \dfrac { du }{ dx } is the derivative of u with respect to x. That is, \dfrac { du }{ dx } =2x.

Thus, once we make the substitution and employ the above relation, we have a new version of the problem to solve: Find F'(x), given F(x)=\int _{ -1 }^{ u }{ -2t+3dt }.

Now, we can apply the Second Fundamental Theorem of Calculus by simply taking the expression { -2t+3dt } and replacing t with x in our solution.

Next, we need to multiply that expression by \dfrac { du }{ dx }. Recall that \dfrac { du }{ dx } =2x, so we will multiply by 2x.

Following these steps gives us our solution: F'(x)=(-2x^{ 2 }+3)(2x)=-4{ x }^{ 3 }+6x.

#### Example 3

Evaluate \dfrac { d }{ dx } \int _{ 0 }^{ x }{ x{ e }^{ -{ t }^{ 2 } } } dt.

As with the examples above, we can evaluate the expression using the Second Fundamental Theorem of Calculus. However, unlike the previous problems, this one includes two variables, x and t. The expression involves a product (two terms being multiplied together), so we must use the product rule.

If you do not remember the product rule, quotient rule, or chain rule, you may wish to go back to these topics and review them at this time. The product rule gives us a method for determining the derivative of the product of two functions. Specifically, it states that for the functions f\left( x \right) and g\left( x \right), the derivative of their product is given by \dfrac { d }{ dx } f(x)g(x)=f(x)g'(x)+g(x)f'(x). In other words, the derivative of the product of two functions is the first function times the derivative of the second plus the second times the derivative of the first.

Let’s apply the product rule to our example. Here, the first function is x, and the second is { e }^{ -{ t }^{ 2 } } . The derivative of x with respect to x is 1, and the derivative of { e }^{ -{ t }^{ 2 } } is \dfrac { d }{ dx } \int _{ 0 }^{ x }{ { e }^{ -{ t }^{ 2 } } }. It is precisely in determining the derivative of this second function that we need to apply the Second Fundamental Theorem of Calculus.

Let’s focus on that now. We are looking for \dfrac { d }{ dx } \int _{ 0 }^{ x }{ { e }^{ -{ t }^{ 2 } } } dt. As you can see, the lower bound is a constant, 0, and the upper bound is x. We can apply the Second Fundamental Theorem of Calculus directly here, and this is a matter of replacing t with x in the expression. Specifically, \dfrac { d }{ dx } \int _{ 0 }^{ x }{ { e }^{ -{ t }^{ 2 } } } dt={ e }^{ -{ x }^{ 2 } }.

Now, let’s return to the entire problem. Applying the product rule, we arrive at the following:

\dfrac { d }{ dx } \int _{ 0 }^{ x }{ x{ e }^{ -{ t }^{ 2 } } } dt=x\int _{ 0 }^{ x }{ { e }^{ -{ t }^{ 2 } } } dt+{ e }^{ -{ t }^{ 2 } }(1)=x{ e }^{ -{ x }^{ 2 } }+{ e }^{ -{ t }^{ 2 } }.

#### Example 4 (Question #15 from the 1998 AP® Calculus Exam)

This multiple choice question from the 1998 exam asked students the following:

If F(x)=\int _{ 0 }^{ x }{ \sqrt { { t }^{ 3 }+1 } dt }, then F'(2) =

The answer choices were as follows:

(a) -3 (b) -2 (c) 2 (d) 3 (e) 18

To solve the problem, we use the Second Fundamental Theorem of Calculus to first find F(x), and then evaluate that function at x=2.

As the lower limit of integration is a constant (0) and the upper limit is x, we can go ahead and apply the theorem directly. Doing so yields F'(x)=\dfrac { d }{ dx } \int _{ 0 }^{ x }{ \sqrt { { t }^{ 3 }+1 } dt } =\sqrt { { x }^{ 3 }+1 }.

Now, we need to evaluate the function we just found for x=2. We substitute 2 for x in the function F’(x), which yields F'(2)=\sqrt { { x }^{ 3 }+1 } =\sqrt { { 2 }^{ 3 }+1 } =\sqrt { 8+1 } =\sqrt { 9 } =3. The solution to the problem is 3, which is choice d.

Example 5 (Free Response Question #3b from the 2012 AP® Calculus Exam)

Part b of this question asks: For each of g'(-3) and g”(-3) find the value or state that it does not exist.

Recall from the question that g(x)=\int _{ 1 }^{ x }{ f(t)dt }. Since we are looking for g'(-3), we must first find g'(x), which is the derivative of the function g with respect to x.

That is, we are looking for g'(x)=\dfrac { d }{ dx } \int _{ 1 }^{ x }{ f(t)dt }. As in previous examples, we can now apply the Second Fundamental Theorem of Calculus. Given that the lower limit of integration is a constant (1) and that the upper limit is x, we can simply replace t with x to obtain our solution. That is, g'(x)=\dfrac { d }{ dx } \int _{ 1 }^{ x }{ f(t)dt } =f(x).

This means that g'(x)=f(x), and g'(-3)=f(-3), which is what we need to find. Thus, we need to find the value of the function f(x) at x=-3.

If we look at the given graph of f(x), we see that at x=-3, the value of the function is 2. Thus, g'(-3)=2. Some of you may not see this easily from the graph. While the graph clearly shows the points (-4, 1) and (-2, 3), it does not explicitly list the coordinates of the point where x=-3. There are a few ways we can go about finding the point on the curve where x=-3.

One way is to determine the slope of the line segment connecting the points (-4, 1) and (-2, 3). The slope is equal to the change in y over the change in x. The change in y is 2 as we move two units up to go from the first point to the second. Meanwhile, the change in x is also two, as we move two units to the right to go from the first point to the second. This makes the slope \dfrac { 2 }{ 2 } =1. If we go back to the point (-4, 1) and use the slope to move one unit up and one unit to the right, we arrive at another point on the segment. This point is (-3, 2), which is the point we are looking for. This point tells us that the value of the function at x=-3 is 2. This is the answer to the first part of the question.

The paragraph above describes the process for finding f(x) in a somewhat intuitive way. If you prefer a more rigorous way, we could also have proceeded as follows. The value of the function at x=-3 is given by the y-coordinate of the point on the curve where x=-3. This point is on the part of the curve that is a line segment. The endpoints of this segment are (-4, 1) and (-2, 3). We can use these to determine the equation of this segment, and from this, the value we seek. Recall that the slope of a line is given by m=\dfrac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } }. Using the points given, we find the slope in this case to be m=\dfrac { 3-1 }{ -2-(-4) } =\dfrac { 2 }{ 2 } =1. Next, we use the slope and one of the endpoints to find the equation of the line segment. We can do this by using the point-slope form of a line:

y-{ y }_{ 1 }=m(x-{ x }_{ 1 }).

Using the point (-4, 1), we obtain y-1=1(x-(-4)). That is, y=x+5. It is important to note that this is the equation of f(x) on the interval [-4, -2]. Now, we can use the equation to find the value of the curve at x=-3. That is, y=-3+5=2, which agrees with our previous solution.

The second part of the question is to find g”(-3). As g'(x)=f(x), g”(x)=f'(x). Thus, we are asked to find the value of the derivative of the function on the graph at x=-3. Since we just found that the equation of the curve on the interval containing x=-3 is y=x+5, the derivative of the function is the slope of this line. The slope of the line is 1 regardless of the value of x.

## Warning

The requirement that f(x) be a continuous function over the interval I containing a is vital. Let’s examine a situation where the function is not continuous over the interval I to see why. As an example, let us consider the function f(x)=\dfrac { 1 }{ x } over the interval [-2, 3], with a=0.

Applying the Second Fundamental Theorem of Calculus with these constraints gives us

F(x)=\int _{ a }^{ t }{ f(t)dt }

F(x)=\int _{ 0 }^{ 3 }{ f(t)dt }

You might be tempted to conclude that F'(x)=f(x), where f(x)=\dfrac { 1 }{ x } and F(x)=\dfrac { { -x }^{ -2 } }{ 2 }.

However, this is not the case, because our original function f(x)=\dfrac { 1 }{ x } is not continuous along the entire interval [-2, 3], as it is not defined for x=0.

Attempting to evaluate the definite integral above makes it clear why the theorem breaks down in this case.

Instead, the First Fundamental Theorem of Calculus gives us the method to evaluate this definite integral. If you do not remember how to evaluate this integral or need to brush up on the First Fundamental Theorem of Calculus, be sure to take a moment to do so.

F(x)={ \left[ \dfrac { 1 }{ x } \right] }_{ 0 }^{ 3 }

F(x)={ \left[ { x }^{ -1 } \right] }_{ 0 }^{ 3 }

F(x)={ \left[ \dfrac { { x }^{ -2 } }{ -2 } \right] }_{ 0 }^{ 3 }

F(x)=\dfrac { { 3 }^{ -2 } }{ -2 } -\dfrac { 0^{ -2 } }{ -2 }

The last fraction is undefined, as it has a zero in the denominator. As such, we cannot determine the value of F(0), which is a direct consequence of trying to apply the theorem incorrectly to a case where the function in question is not continuous over the given interval.

## Summary

The Second Fundamental Theorem of Calculus establishes a relationship between integration and differentiation, the two main concepts in calculus. With this theorem, we can find the derivative of a curve and even evaluate it at certain values of the variable when building an anti-derivative explicitly might not be easy.

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