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AP Calculus AB: Exam Prep21 chapters | 138 lessons | 6 flashcard sets

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Lesson Transcript

Instructor:
*Yuanxin (Amy) Yang Alcocer*

Amy has a master's degree in secondary education and has taught math at a public charter high school.

The fundamental theorem of calculus makes finding your definite integral almost a piece of cake. See how the definite integral becomes a subtraction problem after applying the fundamental theorem of calculus.

Let's get right into this lesson about definite integrals. A **definite integral** is the integral of a function from a starting point to an end point. You'll see little numbers at the top and bottom of your integral sign telling you where your integration begins and where it ends. The bottom value gives you your starting point and the top value gives you your end point.

This is an example of a general, definite integral:

When you read it, you read it saying: 'the definite integral of the function *f* of *x* from point *a* to point *b* with respect to *x*.

To help us evaluate our definite integrals we have a **fundamental theorem of calculus**. This theorem tells us that the definite integral of *f* or *x* from point *a* to point *b*, with respect to *x*, is equal to the integral of *f* of *x* evaluated at point *b* minus the integral of *f* of *x* evaluated at point *a*.

Mathematically we write this where our big *F* stands for the integral. This fundamental theorem of calculus actually helps us a lot when it comes to evaluating our definite integral. It actually makes it very easy for us to do, especially when we've memorized our common integrals as well as our common integration rules, such as the power rule.

Let's take a look at a couple of examples to see how easy it is.

Evaluate:

To evaluate this definite integral, we first find the integral of 3*x*^2, it is 3*x*^3 / 3 = *x*^3 + *C*. We have the constant of integration *C* when we take the general integral. When we take the definite integral we end up subtracting *C* from *C*, which essentially removes it, so we don't worry about it when we work with definite integrals.

Now we can apply the fundamental theorem of calculus. We first evaluate the *x*^3 + *C* at the point *x* = 3. Plugging in 3 for *x* we get: 3^3 + *C* = 27 +*C*.

Now we evaluate the *x*^3 + *C* at the point *x* = 1. Plugging in 1 for *x* we get: 1^3 + *C* = 1 +*C*.

Now subtracting the (1 + *C*) from the (27 + *C*) we get: (27 + *C*) - (1 + *C*) = 27 - 1 = 26.

26 is our answer. Do you see how when we apply the fundamental theorem of calculus the *C* disappears? Know this, we can simply leave out the *C* out all together whenever we work with definite integrals. Just remember to bring the *C* back when you work with general or indefinite integrals.

Evaluate:

We are working with a definite integral so we can leave out the constant of integration throughout our problem. So the integral of 2 cos *x* is 2 sin *x*. Our definite integral then evaluates to 2 sin (*pi* / 2) - 2 sin (0).

We subtract our integral evaluated at 0 from our integral evaluated at (*pi* / 2). Our answer is 2 sin (*pi* / 2) - 2 sin (0) = 2(1) - 2(0) = 2, and we are done.

Since we are given *pi* / 2 we calculate or sign functions using radians. When we see *pi* in our trigonometry angles, it tells us we are working in radians and not degrees.

Let's review what we learned. A **definite integral** is the integral of a function form a starting point to an end point. A definite integral looks like this:

Our start point is *a* and our end point is *b*, and our function is *f*(*x*).

The **fundamental theorem of calculus** tells us that the definite integral of *f* of *x* from point *a* to point *b*, with respect to *x*, is equal to the integral of *f* of *x* evaluated at point *b* minus the integral of *f* of *x* evaluated at point *a*. If we note our integral with a big *F* we can write the fundamental theorem of calculus as this:

To use this fundamental theorem of calculus to help us evaluate definite integrals, we first find our integral. Then we evaluate it at point *b* and point *a*, and then we subtract.

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AP Calculus AB: Exam Prep21 chapters | 138 lessons | 6 flashcard sets

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