# AP Calculus AB : Fundamental Theorem of Calculus

## Example Questions

### Example Question #1 : Use Of The Fundamental Theorem To Evaluate Definite Integrals

Evaluate the following indefinite integral:

Explanation:

Evaluate the following indefinite integral:

Recall that we can split subtraction and addition within integrals into separate integrals. This means that we can look at our problem in two steps.

Recall that we can integrate any exponential term by adding 1 to the exponent and dividing by the new exponent.

So,

Next, recall that the integral of sine is negative cosine. However, we already have a negative sine, so we should get positive cosine.

Now, we can combine our two halves to get our final answer.

Notice that we only have one "c" because c is just a constant, not a variable.

### Example Question #21 : Fundamental Theorem Of Calculus

Evaluate the given definite integral:

Explanation:

Evaluate the given definite integral:

Let's look at this integral as three separate parts, then we'll combine those parts to evaluate the answer.

First, the integral of  is just

Second, The integral of any exponential term can be found by increasing the exponent by 1 and then dividing by the new number.

Third, the integral of cosine is simply positive sine.

Put all this together to get:

Now, we need to evaluate our answer by finding the difference between the limits of integration.

First, let's see what we get when we plug in 0

Next, let's use 5

Next, find the difference of F(5)-F(0)

### Example Question #11 : Use Of The Fundamental Theorem To Evaluate Definite Integrals

Evaluate the following definite integral.

Explanation:

Evaluate the following definite integral.

Lets begin by recalling that taking an integral is essentially the same as reversing the differentiation process.

So, when we integrate sine, we get negative cosine, and when we integrate cosine, we get sine. The coefficients will stay the same.

Now, we have integrated, but we still need to evaluate our integral. To do so, we need to find the difference between  and .

Now, find the difference between our two values:

So, our answer must be 7.

### Example Question #11 : Use Of The Fundamental Theorem To Evaluate Definite Integrals

Explanation:

Derivatives and anti derivatives annihilate each other.  Therefore, the derivative of an anti derivative is simply the function in the integral with the limits substituted in multiplied by the derivative of each limit.  Also, be careful that the units will match the outermost units (which comes from the derivative).

### Example Question #131 : Integrals

Explanation:

We will use the Fundamental Theorem of Calculus

First we find the anti derivative using the rule, then we apply the FTC

Now we just need to simplify,

### Example Question #21 : Fundamental Theorem Of Calculus

Evaluate

Explanation:

Here we use the Fundamental Theorem of Calculus

and the rule

First we find the anti-derivative using the rule above (no constant is needed because we are dealing with definite integrals)

Then we use the FTC by plunging in 3 and subtracting from plugging in 0

### Example Question #12 : Use Of The Fundamental Theorem To Evaluate Definite Integrals

Evaluate

Explanation:

We will use the Fundamental Theorem of Calculus

and the rule

First we find the anti derivative

And then we evaluate, (upper minus lower)

### Example Question #132 : Integrals

Explanation:

We will use the Fundamental Theorem of Calculus

First we find the anti derivative

Then we evaluate it (upper minus lower)

### Example Question #13 : Use Of The Fundamental Theorem To Evaluate Definite Integrals

Find the area of the region given by the following integral.

Explanation:

Find the area of the region given by the following integral.

To evaluate this integral, recall the following rules:

Using these rules, we can find our integral.

Clean it up a bit to get:

Alright, clean it up some more to find our final answer:

So, our area is 3737

### Example Question #11 : Use Of The Fundamental Theorem To Evaluate Definite Integrals

Evaluate the integral

Explanation:

To evaluate a definite integral, that is to say an integral with an upper bound and lower bound, we will use the 2nd Fundamental Theorem of Calculus, which states,

, where is the integral of .

We will first integrate just like with any integration.

We will integrate each term individually.

First, the integral of the cosine is the sine. Doing this, we get

The is just notation for "evaluated from to ". It is simply a reminder to apply the 2nd Fundamental Theorem of Calculus. The  is the from the theorem. Before moving to the second integral, we can apply this theorem.

So far, we have the following expression for the entire problem

The basic integral form for the remaining integral is

Set , since that is what is inside our sine. The derivative of this is

.

This perfectly accounts for the  in in front of the sine. Thus we can immediatly replace the entire integral with its basic integral form's result.

Now we apply the 2nd Fundamental Theorem of Calculus again.

Plug in . Plug in and subtract.