Aim: Fundamental Theorem of Calculus Course: Calculus
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Aim: What is the Fundamental Theorem of Calculus?
Aim: Fundamental Theorem of Calculus Course: 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
Aim: Fundamental Theorem of Calculus Course: Calculus
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
Aim: Fundamental Theorem of Calculus Course: Calculus
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
Aim: Fundamental Theorem of Calculus Course: Calculus
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
Aim: Fundamental Theorem of Calculus Course: Calculus
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
Aim: Fundamental Theorem of Calculus Course: Calculus
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
Aim: Fundamental Theorem of Calculus Course: Calculus
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)
Aim: Fundamental Theorem of Calculus Course: Calculus
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
Aim: Fundamental Theorem of Calculus Course: Calculus
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
Aim: Fundamental Theorem of Calculus Course: Calculus
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]?
Aim: Fundamental Theorem of Calculus Course: Calculus
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 -
Aim: Fundamental Theorem of Calculus Course: Calculus
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
Aim: Fundamental Theorem of Calculus Course: Calculus
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
Aim: Fundamental Theorem of Calculus Course: Calculus
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.
Aim: Fundamental Theorem of Calculus Course: Calculus
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
Aim: Fundamental Theorem of Calculus Course: Calculus
Model Problem
2
0If sin 2 , then '( ) ?
xF x t dt F x
( )x
a
d dFf t dt f x
dx dx
Aim: Fundamental Theorem of Calculus Course: Calculus
Model Problem
5 2 3
21
dt dt
dx
Aim: Fundamental Theorem of Calculus Course: Calculus
Model Problem
21
61
t
dx dx
dt
Aim: Fundamental Theorem of Calculus Course: Calculus
Model Problem2
1 sin
xd tdt
dx t
Aim: Fundamental Theorem of Calculus Course: Calculus
Model Problem
43 2
04
xdt t dt
dx
Aim: Fundamental Theorem of Calculus Course: Calculus
Model Problem
2Find '( ) for ( ) (4 1)
x
xF x F x t dt
Aim: Fundamental Theorem of Calculus Course: Calculus
Model Problem
3 3
0Find '( ) for ( ) 1
xF x F x t dt
Aim: Fundamental Theorem of Calculus Course: Calculus
Model Problem3
2
0Find '( ) for ( ) sin
xF x F x t dt
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