lets take a trip back in time…to geometry. can you find the area of the following? if so, why?

Lets take a trip back in time…to geometry. Can you find the area of the following? If so, why?

Now, lets take a trip back to Advanced Algebra. Can you find the area of the region bounded by the line x=0, y=0 , y = 4 and y = 2x+3? If so, how?

3

04

04

16

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

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.

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. Why?

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 a

xn

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 ank

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.

This will be another day.

We will master the Riemann Sum work first! Onwards!!!

time

velocity

After 4 seconds, the object has gone 12 feet. Why?

Consider an object moving at a constant rate of 3 ft/sec.

Since rate . time = distance:

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

3t d d’ ?

If the velocity is not constant,we might guess that the distance traveled is still equalto the area under the curve.

(The units work out.)

211

8V t Example:

We could estimate the area under the curve by drawing rectangles touching at their left corners.

We call this the Left-hand Rectangular Approx. Method (LRAM).

Approx. area:

We could also use a Right-hand Rectangular Approximation Method(RRAM).

Approx. area:

211

8V t

Another approach would be to use rectangles that touch at the midpoint. This is the Midpoint Rectangular Approximation Method (MRAM).

1.031251.28125

1.78125Approx area: 2.53125

In this example there are four subintervals. As the number of subintervals increases, so does the accuracy.

211

8V t

211

8V t

Approximate area:6.65624

t v

1.007810.25

0.75 1.07031

1.25 1.19531

1.382811.75

2.25

2.75

3.25

3.75

1.63281

1.94531

2.32031

2.75781

13.31248 0.5 6.65624

width of subinterval

With 8 subintervals:

The exact answer for thisproblem is .6.6

Circumscribed rectangles are all above the curve:

Inscribed rectangles are all below the curve:

What Riemann Method?

Over or under estimate?

Concave up or down?

What Riemann Method?

Over or under estimate?

Concave up or down?

Riemann Sums Exercise Handout