19
5.2 Definite Integrals Greg Kelly, Hanford High School, Richland, Washin

5.2 Definite Integrals Greg Kelly, Hanford High School, Richland, Washington

Embed Size (px)

Citation preview

Page 1: 5.2 Definite Integrals Greg Kelly, Hanford High School, Richland, Washington

5.2 Definite Integrals

Greg Kelly, Hanford High School, Richland, Washington

Page 2: 5.2 Definite Integrals Greg Kelly, Hanford High School, Richland, Washington

When we find the area under a curve by adding rectangles, the answer is called a Rieman sum.

211

8V t

subinterval

partition

The width of a rectangle is called a subinterval.

The entire interval is called the partition.

Subintervals do not all have to be the same size.

Page 3: 5.2 Definite Integrals Greg Kelly, Hanford High School, Richland, Washington

211

8V t

subinterval

partition

If the partition is denoted by P, then the length of the longest subinterval is called the norm of P and is denoted by .P

As gets smaller, the approximation for the area gets better.

P

0

1

Area limn

k kP

k

f c x

if P is a partition of the interval ,a b

Page 4: 5.2 Definite Integrals Greg Kelly, Hanford High School, Richland, Washington

0

1

limn

k kP

k

f c x

is called the definite integral of

over .f ,a b

If we use subintervals of equal length, then the length of a

subinterval is:b a

xn

The definite integral is then given by:

1

limn

kn

k

f c x

Page 5: 5.2 Definite Integrals Greg Kelly, Hanford High School, Richland, Washington

1

limn

kn

k

f c x

Leibnitz introduced a simpler notation for the definite integral:

1

limn b

k ank

f c x f x dx

Note that the very small change in x becomes dx.

Page 6: 5.2 Definite Integrals Greg Kelly, Hanford High School, Richland, Washington

b

af x dx

IntegrationSymbol

lower limit of integration

upper limit of integration

integrandvariable of integration

(dummy variable)

It is called a dummy variable because the answer does not depend on the variable chosen.

Page 7: 5.2 Definite Integrals Greg Kelly, Hanford High School, Richland, Washington

b

af x dx

We have the notation for integration, but we still need to learn how to evaluate the integral.

Page 8: 5.2 Definite Integrals Greg Kelly, Hanford High School, Richland, Washington

time

velocity

After 4 seconds, the object has gone 12 feet.

In section 5.1, we considered an object moving at a constant rate of 3 ft/sec.

Since rate . time = distance: 3t d

If we draw a graph of the velocity, the distance that the object travels is equal to the area under the line.

ft3 4 sec 12 ft

sec

Page 9: 5.2 Definite Integrals Greg Kelly, Hanford High School, Richland, Washington

If the velocity varies:

11

2v t

Distance:21

4s t t

(C=0 since s=0 at t=0)

After 4 seconds:1

16 44

s

8s

1Area 1 3 4 8

2

The distance is still equal to the area under the curve!

Notice that the area is a trapezoid.

Page 10: 5.2 Definite Integrals Greg Kelly, Hanford High School, Richland, Washington

211

8v t What if:

We could split the area under the curve into a lot of thin trapezoids, and each trapezoid would behave like the large one in the previous example.

It seems reasonable that the distance will equal the area under the curve.

Page 11: 5.2 Definite Integrals Greg Kelly, Hanford High School, Richland, Washington

211

8

dsv t

dt

31

24s t t

314 4

24s

26

3s

The area under the curve2

63

We can use anti-derivatives to find the area under a curve!

Page 12: 5.2 Definite Integrals Greg Kelly, Hanford High School, Richland, Washington

Let’s look at it another way:

a x

Let area under the

curve from a to x.

(“a” is a constant)

aA x

x h

aA x

Then:

a x aA x A x h A x h

x a aA x h A x h A x

xA x h

aA x h

Page 13: 5.2 Definite Integrals Greg Kelly, Hanford High School, Richland, Washington

x x h

min f max f

The area of a rectangle drawn under the curve would be less than the actual area under the curve.

The area of a rectangle drawn above the curve would be more than the actual area under the curve.

short rectangle area under curve tall rectangle

min max a ah f A x h A x h f

h

min max a aA x h A x

f fh

Page 14: 5.2 Definite Integrals Greg Kelly, Hanford High School, Richland, Washington

min max a aA x h A x

f fh

As h gets smaller, min f and max f get closer together.

0

lim a a

h

A x h A xf x

h

This is the definition

of derivative!

a

dA x f x

dx

Take the anti-derivative of both sides to find an explicit formula for area.

aA x F x c

aA a F a c

0 F a c

F a c initial value

Page 15: 5.2 Definite Integrals Greg Kelly, Hanford High School, Richland, Washington

min max a aA x h A x

f fh

As h gets smaller, min f and max f get closer together.

0

lim a a

h

A x h A xf x

h

a

dA x f x

dx

aA x F x c

aA a F a c

0 F a c

F a c aA x F x F a

Area under curve from a to x = antiderivative at x minus

antiderivative at a.

Page 16: 5.2 Definite Integrals Greg Kelly, Hanford High School, Richland, Washington

0

1

limn

k kP

k

f c x

b

af x dx

F x F a

Area

Page 17: 5.2 Definite Integrals Greg Kelly, Hanford High School, Richland, Washington

Area from x=0to x=1

Example: 2y x

Find the area under the curve from x=1 to x=2.

2 2

1x dx

23

1

1

3x

31 12 1

3 3

8 1

3 3

7

3

Area from x=0to x=2

Area under the curve from x=1 to x=2.

Page 18: 5.2 Definite Integrals Greg Kelly, Hanford High School, Richland, Washington

Example: 2y x

Find the area under the curve from x=1 to x=2.

To do the same problem on the TI-89:

^ 2, ,1,2x x

ENTER

72nd

Page 19: 5.2 Definite Integrals Greg Kelly, Hanford High School, Richland, Washington

Example:

Find the area between the

x-axis and the curve

from to .

cosy x

0x 3

2x

2

3

2

3

2 2

02

cos cos x dx x dx

/ 2 3 / 2

0 / 2sin sinx x

3sin sin 0 sin sin

2 2 2

1 0 1 1 3

On the TI-89:

abs cos , ,0,3 / 2x x 3

If you use the absolute value function, you don’t need to find the roots.

pos.

neg.