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Riemann Sums & Definite Riemann Sums & Definite Integrals Integrals Section 5.3 Section 5.3

Riemann Sums & Definite Integrals Section 5.3. Finding Area with Riemann Sums For convenience, the area of a partition is often divided into subintervals

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Riemann Sums & Definite Riemann Sums & Definite IntegralsIntegralsSection 5.3Section 5.3

Finding Area with Riemann SumsFinding Area with Riemann Sums

• For convenienceFor convenience, the , the area of a partition is area of a partition is often divided into often divided into subintervals with subintervals with equal width – in equal width – in other words, the other words, the rectangles all have rectangles all have the same width. (see the same width. (see the diagram to the the diagram to the right and section 5-2)right and section 5-2)

4

3

2

1

2

f x = x2

Subintervals with equal width

Finding Area with Riemann SumsFinding Area with Riemann Sums

• It is possible to divide a region into It is possible to divide a region into different sized rectangles based on an different sized rectangles based on an algorithm or rule (see graph above algorithm or rule (see graph above and example #1 on page 307)and example #1 on page 307)

6

4

2

5 10 15

Unequal Subintervals

Finding Area with Riemann SumsFinding Area with Riemann Sums

• It is also possible to make rectangles of It is also possible to make rectangles of whatever width you want where the width whatever width you want where the width and/or places where to take the height and/or places where to take the height does not follow any particular pattern.does not follow any particular pattern.

Notice that the subintervals don’t seem to have a pattern. They don’t have to be any specific width or follow any particular pattern.

Also notice that the height can be taken anywhere on each subinterval – not only at endpoints or midpoints!

4

2

What is a Riemann SumWhat is a Riemann SumDefinitionDefinition: :

Let Let f f be defined on the closed interval [a,b], and let be defined on the closed interval [a,b], and let be a partition [a,b] given bybe a partition [a,b] given by

a = xa = x00 < x < x11 < x < x22 < … < x < … < xn-1n-1 < x < xnn = b = b

where where xxii is the width of the is the width of the ith ith subinterval [xsubinterval [xi-1i-1, x, xii]. ]. If c is any point in the If c is any point in the ithith subinterval, then the sum subinterval, then the sum

i-1 i n1

( ) x c x n

i ii

f c x

is called a is called a Riemann sumRiemann sum of of ff for the partition for the partition ..

New Notation for New Notation for xx

• When the partitions (boundaries that tell you where When the partitions (boundaries that tell you where to find the area) are divided into subintervals with to find the area) are divided into subintervals with different widths, the width of the largest subinterval different widths, the width of the largest subinterval of a partition is the NORM of the partition and is of a partition is the NORM of the partition and is denoted by ||denoted by ||||||

• If every subinterval is of equal width, the partition is If every subinterval is of equal width, the partition is REGULAR and the norm is denoted byREGULAR and the norm is denoted by

|| ||||= ||= x = x =

• The number of subintervals in a partition approaches The number of subintervals in a partition approaches infinity as the norm of the partition approaches 0. In infinity as the norm of the partition approaches 0. In other words, ||other words, |||| 0 implies that n || 0 implies that n

b - an

Is the converse of this statement true? Why or why not?

Definite IntegralsDefinite IntegralsIf If f f is defined on the closed interval [a,b] and is defined on the closed interval [a,b] and the limitthe limit

01

lim ( )n

i ii

f c x

exists, then f is exists, then f is integrableintegrable on [a,b] and the on [a,b] and the limit is denoted bylimit is denoted by

The limit is called the The limit is called the definite integraldefinite integral of of ff from from aa to to bb. .

The number The number aa is the is the lower limit of integrationlower limit of integration and the and the number number bb is the is the upper limit of integrationupper limit of integration..

01

( )lim ( )n

i ii

a

bf c f x dxx

Definite Integrals vs. Indefinite Definite Integrals vs. Indefinite IntegralsIntegrals

A A definite integraldefinite integral is is numbernumber..

An An indefinite integralindefinite integral is a is a family of functionsfamily of functions. .

They may look a lot alike, however, They may look a lot alike, however, • definite integrals have limits of integrationdefinite integrals have limits of integration while the while the • indefinite integrals have not limits of indefinite integrals have not limits of integration.integration. ( )

b

af x dx ( )f x dx

Definite Definite IntegralIntegral

Indefinite Indefinite IntegralIntegral

Theorem 5.4Theorem 5.4 Continuity Implies Continuity Implies IntegrabilityIntegrability

If a function If a function ff is is continuous on the closed continuous on the closed interval [interval [a,ba,b], then ], then ff is is integrable on [integrable on [a,ba,b].].

Is the converse of this statement true? Why or why not?

Evaluating a definite integral…Evaluating a definite integral…

To learn how to evaluate a To learn how to evaluate a definite integral as a limit, definite integral as a limit, study Example #2 on p. study Example #2 on p.

310.310.

Theorem 5.5Theorem 5.5 The Definite Integral as the Area of a The Definite Integral as the Area of a RegionRegionIf If ff is is continuouscontinuous and and nonnegativenonnegative on the closed on the closed interval [a,b], then the interval [a,b], then the area of the region bounded area of the region bounded by the by the graph of graph of ff , the , the x-x-axisaxis, and the vertical lines , and the vertical lines x = ax = a and and x = bx = b is given by is given by

( )b

af x dxArea =Area =

5

4

3

2

1

-1

-2 2

f(x) = -0.61x4 + -1.14x3 + 1.75x2 + 2.71x + 2.46(x,f(B))

(x,f(A))

1A B

Let’s try this out…

1. Sketch the region

2. Find the area indicated by the integral.

3

14 dx

8

6

4

2

-2

5Area

= (base)(height)

= (2)(4) = 8 un 2

Give this one a try… 1. Sketch the region

2. Find the area indicated by the integral.

3

0(x+2) dx

8

6

4

2

-2

5

Area of a trapezoid

= .5(width)(base1+ base2)

= (.5)(3)(2+5) = 10.5 un 2

Width =3

base1=2

base2 =5

Try this one… 1. Sketch the region

2. Find the area indicated by the integral.

2 2

24 x dx

2

Area of a semicircle

= .5( r2)

= (.5)()(22)

= 2 un 2

Properties of Definite IntegralsProperties of Definite Integrals

• If If f f is defined at x = a, then we define is defined at x = a, then we define

( ) 0a

af x dx

(x) (x) b

b

a

af dx f dx

So, So,

If If f f is integrable on [a,b], then we define is integrable on [a,b], then we define

So, So,

sin x dx

0

3( 2) x dx

Additive Interval Additive Interval PropertyProperty

(x) (x) (x) a a c

bcbf dx f dx f dx

If If f f is integrable on is integrable on three closed three closed intervals intervals determined by a, b, determined by a, b, and c, then and c, then

Theorem 5.6Theorem 5.6

4

3

2

1

-1

-2

-3

-8 -6 -4 -2 2

0 1 2 3 4 5-1-2-3-4-5

C

To morph f, drag a,b,c,d,e.To rescale sliders, drag red tick mark at 1.

n = 1

f(x) dx = 7.0921.427

Dbl-click n to changeOR select & type +/-

-1.498

f(x) = -0.61x4 + -1.14x3 + 0.75x2 + 2.71x + 2.46

Show RS-LS Info.

n+10 nonstop, n<80

n+1 nonstop, n<80

Hide Integral Shading

Show Trapezoids

Show Left Rects.

Show Right Rects.

Show Mdpt. Rects.(x,f(B))

(x,f(A))

1A B

ab

cd

e

M7

N7

Properties of Properties of Definite Definite IntegralsIntegrals

(x) (x) b b

a akf dx k f dx

If If f and g f and g are integrable on [are integrable on [a,ba,b] and ] and kk is is a constant, then the functions of a constant, then the functions of kf kf and and f f g g are integrable on [ are integrable on [a,ba,b], and], and

Theorem 5.7Theorem 5.7

[ (x) ( )] (x) g(x) b b b

a a af g x dx f dx dx

HOMEWORKHOMEWORK – yep…more – yep…more practicepractice

Thursday, January 17Thursday, January 17

Read and take notes on section 5.3 and Read and take notes on section 5.3 and do p. 314 # 3, 6, 9, …, 45, 47, 49, 53, do p. 314 # 3, 6, 9, …, 45, 47, 49, 53, 65- 70 – Work on the AP Review 65- 70 – Work on the AP Review Diagnostic Tests if you have time over Diagnostic Tests if you have time over the long weekend.the long weekend.

Tuesday – January 22Tuesday – January 22 – –

Get the Riemann sum program for your Get the Riemann sum program for your calculator and do p. 316 # 59-64, 71 calculator and do p. 316 # 59-64, 71 and then READ and TAKE Notes on and then READ and TAKE Notes on section 5-4 and maybe more to come!!section 5-4 and maybe more to come!!