5.2 Definite Integrals

Gre Kelly, Hanford High School, Richland, Washington

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

21 18

V 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.

21 18

V 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

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 ax

n

The definite integral is then given by:

1

limn

kn k

f c x

1

limn

kn k

f c x

Leibnitz introduced a simpler notation for the definite integral:

1

limn b

k an k

f c x f x dx

Note that the very small change in x becomes dx.

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.

b

af x dx

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

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 ftsec

If the velocity varies:

1 12

v 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 82

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

Notice that the area is a trapezoid.

21 18

v 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.

21 18

dsv tdt

3124

s t t

31 4 424

s

263

s

The area under the curve263

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

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

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

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!

ad A x f xdx

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

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

ad A x f xdx

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.

0 1

limn

k kP k

f c x

b

af x dx

F x F a

Area

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

13

x

31 12 13 3

8 13 3

73

Area from x=0to x=2

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

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

Example:

Find the area between the

x-axis and the curve

from to .

cosy x

0x 32

x 2

32

32 2

02

cos cos x dx x dx

/ 2 3 / 2

0 / 2sin sinx x

3sin sin 0 sin sin2 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.

A little check to see if you really got today’s lesson…

• ∫(upper=s, lower =a) ax dx

Yes! Its your Fun and Happy Joy Joy Pleasure Time

#3-27,39, 40,41,43-46Journal: Explain your solution to #27.