7.4: the fundamental theorem of calculus objectives: to use the ftc to evaluate definite integrals...

7.4: The Fundamental Theorem of Calculus

Objectives:•To use the FTC to evaluate definite integrals•To calculate total area under a curve using FTC and geometric formulas•To connect the derivative and the integral using FTC

Properties of the Definite Integral(some repeated from 7.1)

If all indicated definite integrals exist, then:1.

2. for any real # k

a

a

dxxf 0)(

b

a

b

a

dxxfkdxxfk )()(

Properties cont…

3.

4. for any real # c

b

a

b

a

b

a

dxxgdxxfdxxgxf )()()()(

b

a

c

a

b

c

dxxfdxxfdxxf )()()(

Properties cont…

5.

6. If f(x) < g(x) for all x, then

a

b

b

a

dxxfdxxf )()(

b

a

b

a

dxxgdxxf )()(

And last one…

7. , where c is a constant b

a

abccdx )(

Using the Properties

Given:

Evaluate:a.) b.)

c.) d.)

5

0

7

5

3)(;10)( dxxfdxxf

7

0

)( dxxf 0

5

)( dxxf

5

5

)( dxxf 5

0

)(3 dxxf

Evaluate using properties and area:

dxx )135( 21

0

Find if: dxxgxf

2

1

)(2)(

2

1

2

1

3)(;5)( dxxgdxxf

When evaluating the definite integral using area under the curve, subtract areas that fall below the x-axis:

dxx

2

4

2

Fundamental Theorem of Calculus, Part 1

Let

What is A(0), A(1), and A(2)?Find a general formula for A(x) and A’ (x).

x

tdtxA0

)(

Fundamental Theorem of Calculus, Part 1

Connects the integral and the derivative

Suppose f(t) is a continuous function on some interval [a,b]. Let

Then A(x) is differentiable and A’(x) = f(x)(The derivative of the integral function is the

integrand with a change in variable)

x

a

dttfxA )()(

Example

Find y’.

dttyx

0

2 1

Fundamental Theorem of Calculus, Part 2

Involves the antiderivative Shows how to evaluate the definite integral directly

from antiderivatives

Let f be continuous on [a,b], and let F be any antiderivative of f. Then

b

a

baxFaFbFdxxf )()()()(

How to use FTC part 2: (basic problems)

1. Find antiderivative of integrand2. Evaluate the antiderivative for the upper

limit, and subtract the antiderivative of the lower limit

MUST BE CONTINUOUS ON INTERVAL!!FTC2 does have its limitations…not all integrals can be evaluated using it (can’t find antiderivative).

Example:

Evaluate:

4

2

32 dtt

Examples:

1.

2.

3.

dxxx 2

1

23 23

dxx4

1

3

4

0

2 )(sec

dxx

Example:

dxx

x

3

1

2 1

Examples:Evaluate

1.

2.

3.

dxxx

1

3

2 256

dttt 20

4

dww

ww

2

12

5 32

Revisiting FTC part 1…why it works!

1. Find F ‘(x)

2. Find F’(x)

x

dttxF0

)(cos)(

dtttxFx

0

23)(

Can you use FTC part 2 to evaluate the following?

dxx

x

3

2

2

1

1

Using Substitution

Method 1: Integrating in terms of u, need to change upper and lower boundsEvaluate:

dxxx 5

0

225

Evaluate:

dx

x

6

4272

2

Method 2: Find indefinite integral first, then evaluate using original limits

dxxx 5

0

225

Example

The rate at which a substance grows is given by

where x is the time (in days). What is the total accumulated growth during the first 3.5 days?

xexR 2.0150)('

Finding Total Area The definite integral is a number; it accounts for regions of

curve below the x-axis When you use FTC 2, area below the x-axis is subtracted

To find TOTAL AREA between the graph y=f(x) and the x-axis over the interval [a,b]:

1. Partition [a,b] by finding the zeros of f.2. Integrate f over each subinterval3. Add the absolute values of the integrals

Find the area of the region between the curve y=9-x2 and the x-axis over the interval [0,4]

Connecting the ideas…

Graph the function over the interval. a.) Integrate the function over intervalb.) Find the total area of the region between the

graph and the x-axis

y = x2 -6x+8. [0, 3]