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

  =

  200 100

  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.

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 b a 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 1 0 2x dx: First find any antiferivative of 2x, such as F(x) =       Then compute F(1) - F(0), to get         2. To compute 1 -1 (x4 - 3x3 + 1) dx: First find any antiferivative of x4 - 3x3 + 1, such as F(x) =       Then compute F(1) - F(−1), to get         3. To compute 1 0 (et - t) dt: First find any antiferivative of et - t, such as F(t) =       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	=   x5 5 - 3x4 4 + x   1 -1 The square brackets followed by the numbers mean: Put x = the top value, put x = the bottom value and subtract
	= 15 5 - 3(1)4 4 + 1 ↑ We put x = 1 - (-1)5 5 - 3(-1)4 4 + (-1) ↑ We put x = -1
	= 1 5 - 3 4 + 1 + 1 5 + 3 4 + 1 = 12 5 = 2.4

2 0 (2x3 - x2 + 4) dx	=	2 0
	=	-
	=

2 1 x 6 - 6 x dx	=	2 1
	=	-
	=

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

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 + 2 0 x3 dx
	0 -1 x3 dx - 2 0 x3 dx			2 0 x3 dx - 0 -1 x3 dx
	-1 2 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

2 1

x 6

-

6 x

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