## 6.4 The Definite Integral: Algebraic Approach and the Fundamental Theorem of Calculus

(This topic is also in Section 6.4 in Applied Calculus and Section 13.4 in Finite Mathematics and Applied Calculus)

To motivate the Fundamental Theorem of Calculus (see below) let us look at a cost function from the textbook (Example 5 in Section 6.1 of Applied Calculus or Section 13.1 in Finite Mathematics and Applied Calculus):

Example: The marginal cost of producing baseball caps at a production level of x caps is 4 - 0.001x dollars per cap. Find the total change of cost if production is increased from 100 to 200 caps.

We can solve this problem in two ways:

• Using An Antiderivative:
Since the marginal cost is C'(x) = 4 - 0.001x, the total cost is
 C(x)
 =
 C'(x) dx
= (4 - 0.001x) dx   =   4x - 0.0005x2 + K
We don't know the value of the constant K, but we don't really need to; we are asked to find the cost if production is increased from 100 to 200 caps, that is, C(200) - C(100):
 Total Change in Cost = C(200) - C(100) = [4(200) - 0.0005(200)2 + K] - [4(100) - 0.0005(100)2 + K] = [780 + K] - [395 + K] = \$385
so the constant K just cancels out!
Putting aside the actual details of the cost function for now, what this tells us is that:

The total change of cost in going from a items to b is obtained by taking the antiderivative of the marginal cost, evaluating at x = b, evaluating at x = a, and then subtracting the answers.

But Wait! we also have this second way of doing this calculation based on the method in the previous tutorial:

• Using a Definite Integral:
Since C'(x) is the rate of change of the cost function, the last tutorial tells us that the total change in C from x = 100 to x = 200 is
 Total Change in Cost = 200100 C'(x) dx

This calculation says:

The total change of cost in going from a items to b is obtained by taking the definite integral of the marginal cost from a to b.

Comparing these two methods of computing the total change, we see that the first method gives us a way of calculating the definite integral in the second method without having to calculate Riemann sums! This is the central idea of the Fundamental Theorem of Calculus:

Fundamental Theorem of Calculus (FTC)

Let f be a continuous function defined on the interval [a, b] and if F is any antiderivative of f and is defined on [a, b], we have

 ba f(x) dx = F(b) - F(a)

Moreover, such an antiderivative is guaranteed to exist.

In Words
Every continuous function has an antiderivative. To compute the definite integral of f(x) over [a, b], first find an antiderivative F(x), then evaluate it at x = b, evaluate it at x = a, and subtract the two answers.

Examples

 1 To compute 10 2x dx:
1. First find any antiferivative of 2x, such as F(x) =
2. Then compute F(1) - F(0), to get
 2 To compute 1-1 (x4 - 3x3 + 1) dx:
1. First find any antiferivative of x4 - 3x3 + 1, such as F(x) =
2. Then compute F(1) - F(−1), to get
 3 To compute 10 (et - t) dt:
1. First find any antiferivative of et - t, such as F(t) =
2. Then compute F(1) - F(0), to get

Notation: Let us redo Example 2 above, but this time introduce some notation as we go:

 1-1 (x4 - 3x3 + 1) dx
 = x55 - 3x44 + x 1-1 The square brackets followed by the numbers mean:Put x = the top value, put x = the bottom value and subtract
 = 155 - 3(1)44 + 1
 ↑ We put x = 1
 - (-1)55 - 3(-1)44 + (-1)
 ↑ We put x = -1
 = 15 - 34 + 1 + 15 + 34 + 1 = 125 = 2.4

Q Fill in the correct formulas and values (use C = 0 as your arbitrary constant):
 20 (2x3 - x2 + 4) dx
=
 20
=
 -
=
Q Repeat for the following (again use C = 0 as your arbitrary constant):
 21 x6 - 6x dx
=
 21
=
 -
=

If we remember that definite integrals are also areas, we can use the FTC to compute areas of regions bounded by curves:

In this practice exercise, we will user the Fundamental Theorem of Calculus find the area bounded by various lines and curves in the xy-plane.

Q Consider the area in the xy-plane bounded by the x-axis, the vertial lines x = -1 and x = 2, and the curve y = x3:

To compute the total area (red + green) shown, we should compute:

 2-1 x3 dx
 0-1 x3 dx + 20 x3 dx
 0-1 x3 dx - 20 x3 dx
 20 x3 dx - 0-1 x3 dx
 -12 x3 dx
 2-1 x3 dx
Get me out of here!

It is often necessary to use substitution to evaluate definite integrals. When doing so for a definite integral, remember that the limits of integration are vvalues of the variable of integraion. For instance, in

 21 x6 - 6x dx

the variable of integration is x, and so the limits (1 and 2) are values of x as well. When we change variables, we should change these values to the corresponing values of the new variable as we now show:

Using the FTC with Substitution

When you change variables from, say, x to u, you need to remember that the limits in the given definite integral are limits of x, and not u. So it is a good idea to change the limits to values of u, as we see in the following example:

Example

 To compute 10 x + 1ex2 + 2x-4 dx
1. Decide what substitution to use: Set u =
2. Find dx in terms of du: We get du =
3. Next, compute the values for u that correspond to the limits of integration, which are currently values of x:
 x = 1x =0 x + 1ex2 + 2x-4 dx
The corresponding values of u are:
 x = 0 (bottom limit) ⇒ u = x = 1 (top limit) ⇒ u =
4. After making the substitution, the transformed integral is:
 10 x + 1ex2 + 2x-4 dx =
 10 e-u du
 2 10 e-u du
 12 10 e-u du
 -1-4 e-u du
 2 -1-4 e-u du
 12 -1-4 e-u du
Get me out of here!
5. Finally, we evaluate the correct integral above to get:

You now have several options

• Try some of the questions in the true/false quiz by pressing the button on the sidebar.
• Try the on-line review exercises (press the "Review Exercises" button on the side.
• Try the exercises from Section 6.4 in Applied Calculus or Section or Section 13.4 of Finite Mathematics and Applied Calculus.

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Last Updated: November, 2006