Aim: Fundamental Theorem of Calculus Course: Calculus Do Now: Aim: What is the Fundamental Theorem of Calculus?

Aim: Fundamental Theorem of Calculus Course: Calculus

Do Now:

Aim: What is the Fundamental Theorem of Calculus?


Inverses of each other

Connection: Differentiation & Integration

Two branches of calculus:

Differentiation - rate of change

Integration – accretion (area)

Δy

Δx

Slope = y

x

seca

nt

y

x

tang

ent

Δy

Δx

Area = y x

area of rectangle

area of region

y x precalculus precalculus


Connection: Differentiation & Integration

Calculus is the study of limits

0

( )limh

f x h f x

h

derivative of a function

01

( ) lim ( )nb

i iai

f x dx f c x

definite integral

Fundamental Theorem of Arithmeticwhole numbers can be factored into product of primes

two most important

limits

Fundamental Theorem of Algebranth degree polynomial has n roots


The Fundamental Theorem of Calculus

If a function of f is continuous on the closed interval [a, b] and F is an antiderivative of f on the interval [a, b], then

( ) ( )b

af x dx F b F a

Guidelines1. You now have a way to evaluate a

definite integral without using the limit of a sum.

2. Use the following notation

3. It is not necessary to use the constant of integration C

( )b b

aaf x dx F x


Evaluating a Definite Integral

2 2

13x dx

4

13 xdx

4 2

0sec x dx

23

1

33

xx

8 1 26 3

3 3 3

43 24 1 2

11

3 33 2

xx dx

3 2 3 22 4 2 1 14

4

0tan 1 0 1x

( )b b

aaf x dx F x


Evaluating an Absolute Value

2

02 1x dx

1 2 22 2

0 1 2x x x x

1 1 1 10 0 4 2

4 2 4 2

1(2 1),

22 11

2 1, 2

x xx

x x

2

02 1x dx

1 2

0(2 1)x dx

5

2

4

3

2

1

2

2

1 2(2 1)x dx


Finding Area of Region

Find the area of the region bounded by the graph of y = 2x2 – 3x + 2, the x-axis, and the vertical lines x = 0 and x = 2.

5

4

3

2

1

2

2 2

0Area = 2 3 2x x dx

Integrate [0, 2]

23 2

0

2 32

3 2

x xx

Find F(x)

16 106 4 0 0 0

3 3

Evaluate F(2) – F(0)

( )b b

aaf x dx F x


Mean Value Theorem of Integrals

If f is continuous on the closed interval [a, b], then there exists a number c in the closed interval [a, b] such that

( )( )b

af x dx f c b a

a b

f

c

somewhere between the inscribed and circumscribed rectangles there is a rectangle whose area is equal to the area of the region

under the curve.

= average value of f on [a, b]

f(c)


Average Value of a Function

If f is integrable on the closed interval [a, b], then the average value of f on the interval is

1 b

af x dx

b a

Find the average value of f(x) = 3x2 – 2x on the interval [1, 4]

4 2

1

13 2

4 1x x dx

43 2

1

1

3x x 1 48

64 16 (1 1) 163 3


Average Value of a Function

Find the average value of f(x) = 3x2 – 2x on the interval [1, 4]

4 2

1

13 2

4 1x x dx

16

40

35

30

25

20

15

10

5

1 2 3 4

f x = 3x2-2x

f(c) 16

3x2 – 2x = 16

average value

c = 8/3

c


Model Problem

At different altitudes in earth’s atmosphere, sound travels at different speeds. The speed of sound s(x) (in meters per second) can be modeled by

where x is the altitude in kilometers. What is the average speed of sound over the interval [0, 80]?


Model Problem

11.5

0s x dx

11.5

04 341 3657x dx

22

11.5s x dx

22

11.5295 3097.5dx

32

22s x dx

32

22

3278.5 2987.5

4x dx

50

32s x dx

50

32

3254.5 5688

2x dx

80

50s x dx

80

50

3404.5 9210

2x dx

80

024,640s x dx

80

0

1 24,640308 meters per second

80 80s x dx

Sum of 5 integrals -


new variable of integration

Accumulation Function

b

af x dx

Definite Integral as a Number

Constant

Constant

f is a function

of x

( )x

aF x f t dt

Definite Integral as a Function of x

Constant

F is a function of x

f is a function

of t

Accumulation function: area accumulates under a curve from fixed value of (t = a) to a variable value (t = x)

Note: definite integral is not function of variable of integration


Model Problem

Evaluate0

( ) cos

at 0, 6, 4, 3, and 2.

xF x t dt

x

6 4 3 2

06 4 2

cos cos cos cost dt t dt t dt t dt

option 1

00cos sin

x xt dt t

option 2

sin sin 0x sin ( )x F x

( ) sin(0) 0F x

1( ) sin( )

6 6 2F

2( ) sin( )

4 4 2F

3( ) sin( )

3 3 2F

( ) sin( ) 12 2

F


2nd Fundamental Theorem of Calculus

( )x

a

d dFf t dt f x

dx dx

If f is continuous on an open interval I containing a, then, for

every x in the interval,

Evaluate 2

01

xdt dt

dx 2 1x

Caution: if the upper limit is a function of x, ex. x2, then the answer is multiplied by the derivative of the upper limit term.


Model Problem

Find the derivative of F(x)

3

2cos

xF x t dt

3u x '( )dF du

F xdu dx

chain rule

2cos

ud dut dt

du dx

2

3 2

(cos )(3 )

(cos )(3 )

u x

x x

3 3

3

22cos sin (sin ) 1

x xF x t dt t x

verification

3 3 2sin 1 cos 3d

x x xdx

( )x

a

d dFf t dt f x

dx dx


Model Problem

2

0If sin 2 , then '( ) ?

xF x t dt F x

( )x

a

d dFf t dt f x

dx dx


Model Problem

5 2 3

21

dt dt

dx


Model Problem

21

61

t

dx dx

dt


Model Problem2

1 sin

xd tdt

dx t


Model Problem

43 2

04

xdt t dt

dx


Model Problem

2Find '( ) for ( ) (4 1)

x

xF x F x t dt


Model Problem

3 3

0Find '( ) for ( ) 1

xF x F x t dt


Model Problem3

2

0Find '( ) for ( ) sin

xF x F x t dt

Aim: Fundamental Theorem of Calculus Course: Calculus Do Now: Aim: What is the Fundamental Theorem of Calculus?

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