5.2 Definite Integrals

Bernhard Reimann

30

34

kk=2

50

∑2k

k=2

50

∑3k 2

k=2

100

∑

n2 + 3nk=1

4

∑0

1

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

0

1

2

3

1 2 3 4

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.

→

0

1

2

3

1 2 3 4

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→

=

= Δ∑ if P is a partition of the interval [ ],a b

→

( )0

1

limn

k kP

k

f c x→

=

Δ∑ 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

−Δ =

The definite integral is then given by:

( )1

limn

kn

k

f c x→∞

=

Δ∑

→

( )1

limn

kn

k

f c x→∞

=

Δ∑ Leibnitz introduced a simpler notation for the definite integral:

( ) ( )1

limn b

k ank

f c x f x dx→∞

=

Δ =∑ ∫

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 it is integrated out in the final answer. →

( )b

af x dx∫

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

→

0

1

2

3

1 2 3 4

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.

→

S 4( )= 3 dt 04∫

= 12

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.

→

0

1

2

3

1 2 3 4x

211

8

dsv t

dt= = +

31

24s t t= +

s =

124

43 +4

26

3s =

The area under the curve2

63

=

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

→

0

1

2

3

1 2 3 4x

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!

( ) ( )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

→

( ) ( )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− = A

ax( ) =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→

=

= Δ∑

( )b

af x dx= ∫

=F b( )−F a( )

Area

→

f x( )a

b∫ dx = fnInt f(x),x,a,b( )

On your Ti-84 calculator:

Area from x=0to x=1

0

1

2

3

4

1 2

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.

→

-1

0

1

Example:

Find the area between the

x-axis and the curve

from to .

cosy x=

0x = 3

2x

π= 2

π

3

2

π

3

2 2

02

cos cos x dx x dxπ π

π−∫ ∫/ 2 3 / 2

0 / 2sin sinx x

π π

π−

3sin sin 0 sin sin

2 2 2

π π π⎛ ⎞ ⎛ ⎞− − −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

( ) ( )1 0 1 1− − − − 3=

On the TI-89:

( )( )( )abs cos , ,0,3 / 2x x π∫3=

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

+

-

π

-1

0

1

Example:

Find the area between the

x-axis and the curve

from to .

cosy x=

0x = 3

2x

π= 2

π

3

2

π

3

2 2

02

cos cos x dx x dxπ π

π−∫ ∫/ 2 3 / 2

0 / 2sin sinx x

π π

π−

3sin sin 0 sin sin

2 2 2

π π π⎛ ⎞ ⎛ ⎞− − −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

( ) ( )1 0 1 1− − − − 3=

On the Ti-84:

3=

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

+

-

fnInt abs cos x( )( ),x,0,3π / 2( )

Example2 2

2Evaluate the integral 4 - x dx

−∫

-2 2

3

2

1

-1

A =12π 2( )2 =2π

Example

Evaluate the integral sin x dxπ

−π∫2

1

-1

-2

-2 2

Example

( ) ( ) ( )b

a

Evaluate the integral sin x dx = 0

Note : f x dx = area above axis - area below axis

π

−π∫

∫2

1

-1

-2

-2 2

A =cos(x) −ππ

=cos π( )−cos −π( ) =0

Discontinuous Functions

Some functions with discontinuities are also integrable. A bounded function that has a finite number of points of discontinuity on an interval [a, b] will still be integrable on the interval if it is continuous everywhere else.

2

1

xFind dx

x−∫

Discontinuous Function

2

1

x dx

x

= 1 1 + 1 2

= 1

−

⋅− ⋅

∫

Try this on your calculator using fnInt…

2

1

-1

-2

-2 2

f x( ) = x

x