Riemann Sums and the Definite Integral

Lesson 5.3

Why?

• Why is the area of the yellow rectangle at the end =

a bx

( ) ( )x f b f a

Review

• We partition the interval into n sub-intervals

• Evaluate f(x) at right endpointsof kth sub-interval for k = 1, 2, 3, … n

a b

f(x)

b ax

n

a k x

Review

• Sum

• We expect Sn to improve thus we define A, the area under the curve, to equal the above limit.

a b

1

lim ( )n

nn

k

S f a k x x

f(x)

Look at Goegebra demo

Look at Goegebra demo

Riemann Sum

1. Partition the interval [a,b] into n subintervalsa = x0 < x1 … < xn-1< xn = b

• Call this partition P

• The kth subinterval is xk = xk-1 – xk

• Largest xk is called the norm, called ||P||

2. Choose an arbitrary value from each

subinterval, call it ic

Riemann Sum3. Form the sum

This is the Riemann sum associated with• the function f• the given partition P• the chosen subinterval representatives

• We will express a variety of quantities in terms of the Riemann sum

1 1 2 21

( ) ( ) ... ( ) ( )n

n n n i ii

R f c x f c x f c x f c x

1 1 2 21

( ) ( ) ... ( ) ( )n

n n n i ii

R f c x f c x f c x f c x

ic

The Riemann SumCalculated

• Consider the function2x2 – 7x + 5

• Use x = 0.1

• Let the = left edgeof each subinterval

• Note the sum

x 2x 2̂-7x+5 dx * f(x)4 9 0.9

4.1 9.92 0.9924.2 10.88 1.0884.3 11.88 1.1884.4 12.92 1.2924.5 14 1.44.6 15.12 1.5124.7 16.28 1.6284.8 17.48 1.7484.9 18.72 1.872

5 20 25.1 21.32 2.1325.2 22.68 2.2685.3 24.08 2.4085.4 25.52 2.5525.5 27 2.75.6 28.52 2.8525.7 30.08 3.0085.8 31.68 3.1685.9 33.32 3.332

Riemann sum = 40.04

x 2x 2̂-7x+5 dx * f(x)4 9 0.9

4.1 9.92 0.9924.2 10.88 1.0884.3 11.88 1.1884.4 12.92 1.2924.5 14 1.44.6 15.12 1.5124.7 16.28 1.6284.8 17.48 1.7484.9 18.72 1.872

5 20 25.1 21.32 2.1325.2 22.68 2.2685.3 24.08 2.4085.4 25.52 2.5525.5 27 2.75.6 28.52 2.8525.7 30.08 3.0085.8 31.68 3.1685.9 33.32 3.332

Riemann sum = 40.04

ic

The Riemann Sum

• We have summed a series of boxes

• If the x were smaller, we would have gotten a better approximation

f(x) = 2x2 – 7x + 5

1

( ) 40.04n

i ii

f c x

The Definite Integral

• The definite integral is the limit of the Riemann sum

• We say that f is integrable when the number I can be approximated as accurate

as needed by making ||P|| sufficiently small f must exist on [a,b] and the Riemann sum

must exist

0

1

lim( )b

a P

n

i ik

f f c xI x dx

Example

• Try

• Use summation on calculator.

3 4

24

11

use (1 )k

x dx S f k x x

b ax

n

Example

• Note increased accuracy with smaller x

Limit of the Riemann Sum

• The definite integral is the limit of the Riemann sum.

3

2

1

x dx

Properties of Definite Integral

• Integral of a sum = sum of integrals

• Factor out a constant

• Dominance

( ) ( ) [ , ]

( ) ( )b b

a a

f x g x on a b

f x dx g x dx

Properties of Definite Integral

• Subdivision rule

( ) ( ) ( )c b c

a a b

f x dx f x dx f x dx

a b c

f(x)

Area As An Integral

• The area under the curve on theinterval [a,b] a c

f(x)

( )b

a

A f x dx

A

Distance As An Integral

• Given that v(t) = the velocity function with respect to time:

• Then Distance traveled can be determined by a definite integral

• Think of a summation for many small time slices of distance

( )t b

t a

D v t dt

Assignment

• Section 5.3

• Page 314

• Problems: 3 – 47 odd