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If the partition is denoted by P, then the length of the longest subinterval is called the norm of P and is denoted by . As gets smaller, the approximation for the area gets better. if P is a partition of the interval Subinterval (partition) P P 0 1 A rea lim n k k P k f c x , ab 2 1 1 8 V t

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

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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

211

8V t

Subinterval (partition)

Learning Target: The definite integral is limit of the Riemann SumIntegral measures accumulated change.

4.3 Definite Integrals

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

211

8V t

Subinterval (partition)

The width of a rectangle is called a subinterval.

Subintervals do not all have to be the same size.

The Definite Integral

Leibnitz introduced this simpler notation to represent the Rimeann Sum

0

1

limn b

k aPk

f c x f x dx

Note that the very small change in x becomes dx.

The limit of the Riemann Sum, asPartition width goes to zeo

b

af x dx

IntegrationSymbol

lower limit of integration

upper limit of integration

integrandvariable of integration

1 2 3 4

1

2

3

The Definite Integral as the Area of a Region: If f is continuous and nonnegative on [a, b], then the area of the region bounded by the graph of f, the x-axis and x = a and x = b is …

( )

b

a

Area f x dx

1 2 3 4

1

2

3

• Given: • Write the integral that

represents the area under the curve from 0 to 4.

1. Set up a definite integral that yields the area of the region.2. Use a geometric formula to evaluate the integral.

( ) 5f x x

-6 -5 -4 -3 -2 -1 1 2 3 4 5 6

-2

-1

1

2

3

4

5

6

7

8

9

10

“Negative” area

Regions below the x-axis have “negative” area.

Why?

p. 274 #47.

Properties of Definite Integrals:

2. 0a

af x dx If the upper and lower limits are equal,

then the integral is zero.

1. b a

a bf x dx f x dx Reversing the limits

changes the sign.

b b

a ak f x dx k f x dx 3. Constant multiples can be

moved outside.

1.

0a

af x dx If the upper and lower limits are equal,

then the integral is zero.2.

b a

a bf x dx f x dx Reversing the limits

changes the sign.

b b

a ak f x dx k f x dx 3. Constant multiples can be

moved outside.

b b b

a a af x g x dx f x dx g x dx 4.

Integrals can be added and subtracted.

b b b

a a af x g x dx f x dx g x dx 4.

Integrals can be added and subtracted.

5. b c c

a b af x dx f x dx f x dx

Intervals can be added(or subtracted.)

a b c

y f x

Evaluate the integral using the following values:

5 8 5 54 4 2

1 5 1 1

20 8 12 6x dx x dx x dx dx

54 2

1

84

1

54

8

54

1

1) (2 3 4)

2)

3)

4) (7 3 )

x x dx

x dx

x dx

x dx

4.3 Definite IntegralsIf I have ever made any valuable discoveries, it has been owing

more to patient attention, than to any other talent.

Homework: p. 273 #17-23(odd), 29-43(odd)