section 4.4 – the fundamental theorem of calculus
DESCRIPTION
Section 4.4 – The Fundamental Theorem of Calculus. The First Fundamental Theorem of Calculus. If f is continuous on the interval [ a , b ] and F is any function that satisfies F '( x ) = f ( x ) throughout this interval then. Alternative forms:. Example 1. Evaluate . - PowerPoint PPT PresentationTRANSCRIPT
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Section 4.4 – The Fundamental Theorem of Calculus
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The First Fundamental Theorem of Calculus
If f is continuous on the interval [a,b] and F is any function that satisfies F '(x) = f(x) throughout this interval then
b
af x dx F b F a
Alternative forms: '
b
af x dx f b f a
b
aF b F a f x dx
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Example 1Evaluate
First Find the indefinte integral
F(x): 2F x x dx 2 11
2 1 x C 31
3 x C
13
Now apply the FTC to find the definite
integral:
F b F a 1 0F F
3 31 13 31 0C C
13 C C Notice that it is
not necessary to include the “C”
with definite integrals
1 2
0x dx
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Example 2Evaluate
First Find the indefinte integral
F(x): 3 2F x x dx 3 11
3 1 2x x C
414 2x x C
394
Now apply the FTC to find the definite
integral:
F b F a 2 1F F
4 41 14 42 2 2 1 2 1C C
748 C C Notice that it is
not necessary to include the “C”
with definite integrals
2 3
12x dx
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2
2cos x dx
More Examples: New Notation4. Evaluate
2
2sin x
2 2sin sin 1 1 F(x) Bounds
2
5
103 xx x dx
5. Evaluate
5313 3
10x x
331 13 35 10 5 3 10 3 25
3 21 883
91
4 xx dx6. Evaluate
91 2 212 4
2x x 1 2 2 1 2 21 12 22 9 9 2 4 4
34.5
9 1 2
4x x dx
5 2
310x dx
If needed, rewrite.
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Example 7Calculate the total area between the curve y = 1 – x2 and the
x-axis over the interval [0,2].The question considers all
area to be positive (not signed area), thus use the absolute value function:
21 x
2
2
2
1 , 1
1 , 1 1
1 , 1
x x
x x
x x
2
Use a integral and the piece-wise function to find the area:2 2
01 x dx 1 22 2
0 11 1x dx x dx
3 31 2
3 30 1
x xx x
33 3 301 2 13 3 3 31 0 2 1
Rewrite the equation as a
piece-wise function.
2 43 3
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Example 8Assume F '(x) = f (x), f (x) = sin (x2), and F(2) = -5. Find
F (1).Use the First Fundamental
Theorem of Calculus: 2 2
1sin 2 1x dx F F
2 2
1sin 5 1x dx F
We do not have the ability to analytically
calculate this integral. It will either be given
or you can use a calculator to evaluate
the integral.
0.495 5 1F
1 5.495F
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Example 9The graph below is of the function f '(x).
If f (4) = 3, find f (12).Use the First Fundamental
Theorem of Calculus: 12
4' 12 4f x dx f f
212 4 12 3f
12 3 8f
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White Board ChallengeIf, for all x, f '(x) = (x – 2)4(x – 1)3, it follows that the function f has:
a) a relative minimum at x = 1.b) a relative maximum at x = 1.c) both a relative minimum at x = 1 and a
relative maximum at x = 2.d) neither a relative maximum nor a relative
minimum.e) relative minima at x = 1 and at x = 2.
Multiple C
hoice
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F b F ab af c
Let F be a function that satisfies the following hypotheses:1. F is continuous on the closed interval [a,b]2. F is differentiable on the open interval (a,b)Then there is a number c in (a,b) such that:
Mean Value TheoremLet f be a function that satisfies the following hypotheses:1. f is continuous on the closed interval [a,b]2. f is differentiable on the open interval (a,b)Then there is a number c in (a,b) such that:
' f b f ab af c
b
af x dx
b af c
b
ab a f c f x dx
Redefine the Conditions
Rewrite with integral notation.
Solve for the integral.
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Mean Value Theorem for Integrals
If f(x) is continuous on [a,b], then there exists a value c on the interval [a,b] such that:
b
af x dx b a f c
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Average Value of a FunctionThe average value of an integrable function f(x) on [a,b] is the quantity:
1 b
af x dx
b a
This is also referred to as the Mean Value and can be
described as the average height of a graph.
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Reminder: Average Rate of Change
For a ≠ b, the average rate of change of f over time [a,b] is the ratio:
f b f ab a
Approximates the derivative of a function.
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Example 1Find the average value of f (x) = sin x on [0,π].
Use the Formula:
0
1 sin0
x dx
0
1 cos x
1 cos cos0
1 1 1
2
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Example 2The height of a jump of a bushbaby is modeled by h (t)
= v0t – ½gt2. If g = 980 cm/s2 and the initial velocity is v
0=600
cm/s, find the average speed during the jump.
Use the Average Value Formula: 1 b
af x dx
b a
210 2h t v t gt
212600 980h t t t
2600 490h t t t
We are trying to find the average value of SPEED (absolute value of velocity).
So we need to find the velocity function.
Velocity is the derivative of Position.
'v t h tThus, the speed function is:
600 980t
Now find when the jump begins and ends (a and b).
20 600 490t t 60490,t
Evaluate the integral:6049
60 049
1 600 9800
t dt
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Example 2 (Continued)
Rewrite the equation as a piece-wise function: 600 980t
3049
30 6049 49
600 980 , 0600 980 ,
t tt t
300 /cm s
Use a integral and the piece-wise function to find the average value:6049
60 049
1 600 980t dt 30 6049 49
3049
4960 0
600 980 600 980t dt t dt
30 6049 49
3049
2 24960 0
600 490 600 490t t t t 2 2 2249 30 30 60 60 30 30
60 49 49 49 49 49 49600 490 600 0 490 0 600 490 600 490 49 1800060 49
The height of a jump of a bushbaby is modeled by h (t) = v
0t – ½gt2. If g = 980 cm/s2 and the initial velocity is v
0=600
cm/s, find the average speed during the jump.
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White Board Challenge
Evaluate
2
0
coslimh
hh
1
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Net Change of a Quantity over a Specified Interval
Consider the following problems:1. Water flows into an empty bucket at a rate of
1.5 liters/second. How much water is in the bucket after 4 seconds?
2. Suppose the flow rate varies with time and can be represented as r(t). How much water is in the bucket after 4 seconds?
Quantity of water flow rate time elapsed 1.5 4 6 liters
The quantity of water is equal to the area under the curve of r(t)
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Net Change of a Quantity over a Specified Interval
Water flows into an empty bucket at a rate of 1.5 liters/second. Suppose the flow rate varies with time and can be represented as r(t). How much water is in the bucket after 4 seconds?
4
0' 4 0s t dt s s
The quantity of water is equal to the area under the curve of r(t). Let s(t) be the amount of water in the bucket at time t.
Use the First Fundamental Theorem of Calculus:
4
04 0r t dt s
4
04s r t dt
Signed Area under the graphWater in the bucket at 4 s
IMPORTANT:If the bucket did not
start empty, the integral would
represent the net change of water.
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Net Change as the Integral of a Rate
The net change in s(t) over an interval [t1,t
2]
is given by the integral:
2
12 1'
t
ts t dt s t s t
Integral of the rate of change
Net change from t1 to t
2
Rate at which s(t) is changing
Amount of the quantity at t
1
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Example 1If b(t) is the rate of growth of the number of bacteria in a dish measured in number of bacteria per hour, what does the following integral represent? Be specific.
c
ab t dt
The increase in the number of bacteria from hour a to hour c.
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Example 2The number of cars per hour passing an observation point along a highway is called the traffic flow rate q(t) (in cars per hour). The flow rate is recorded in the table below. Estimate the number of cars using the highway during this 2-hour period.
t 7:00 7:15 7:30 7:45 8:00 8:15 8:30 8:45 9:00
q(t) 1044 1297 1478 1844 1451 1378 1155 802 542
Since there is no function, we can not use the First Fundamental Theorem of Calculus. Instead approximate the area under the curve with any
Riemann Sum (I will use right-endpoints with 0.25 hour lengths):
9:00
7:00q t dt
0.25 1297 1478 1844 1451 1378 1155 802 542
2550 cars
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Example 3A particle has velocity v(t) = t3 – 10t2 + 24 t. Without evaluating, write an integral that represents the following quantities:a) Displacement over [0,6]
b) Total distance traveled over [3,5]
6 3 2
010 24t t t dt
5 3 2
310 24t t t dt
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The Integral of VelocityAssume an object is in linear motion s(t) with velocity v(t). Since v(t) = s'(t):
2
1
2
1
1 2
1 2
Displacement during ,
Distance traveled during ,
t
t
t
t
t t v t dt
t t v t dt
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White Board ChallengeA factory produces bicycles at a rate of:
bicycles per week. How many bicycles were produced from the beginning of week 2 to the end of week 3?
295 3p t t t
3
1212p t dt Bicycles
Week 1 Week 2 Week 3 Week 40 1 2 3 4
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a x b
Signed Area = g(x)
The Definite Integral as a Function of x
Let f be a continuous function on [a, b] and x varies between a and b. If x varies, the following is a function of x denoted by g(x):
x
ag x f t dt
Notice, g(x) satisfies the initial condition g(a) = 0.
Notice that a is a real number.
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Example 1Use the function F(x) to answer the questions below:
3
1
xF x t dt
a) Find a formula for the function.
b) Evaluate F(4).
c) Find the derivative of F(x).
3
1
xF x t dt 41
4 1
xt 4 41 1
4 4 1x 41 14 4x
4F 41 14 44 63.75
'F x 4 114 4 0x 3x Notice that this is the same
as the integral when t = x.
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Example 2Evaluate:
3
21
xd t dtdx
In the previous example, in order to find the derivative we
had to find the integral:
3
21
xt dt
Unfortunately, like many integrals, we can not find an antiderivative for this function.
It should be clear there is an inverse relationship between the derivative and the integral. Thus, the derivative of the
integral function is simply the original function.
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The Second Fundamental Theorem of Calculus
Assume that f(x) is continuous on an open interval I containing a. Then the area function:
x
a
d f t dt f xdx
x
aA x f t dt
is an antiderivative of f(x) on I; that is, A'(x) = f(x). Equivalently,
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Example 2 (Continued)Evaluate:
3
21
xd t dtdx
Since 31f t t
3 3
21 1
xd t dt xdx
f(x)
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Example 3Evaluate:
2
2sin
xd t dtdx
Notice the upper limit of the integral is a function of x rather than x itself. We can not apply the 2nd FTC. But
we can find an antiderivative of the integral:2
2sin
xt dt
2
2cos
xt
2cos cos 2x
2cos cos 2x Find the derivative of the result:
2cos cos 2d xdx
2sin 2 0x x 22 sinx x
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Example 4Evaluate:
13
1cosxd t dt
dx
Notice we can not find an antiderivative of the integral AND the upper limit of the integral is a function of x
rather than x itself.
How do we handle this? Can we apply the 2nd FTC?
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The Upper Limit of the Integral is a Function of x
Use the First Fundamental Theorem of Calculus to evaluate the integral:
d F g x F adx
g x
af t dt
Find the derivative of the result:
F g x F a
d dF g x F adx dx
' ' 0F g x g x
'f g x g x
Chain Rule Constant
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Composite Functions and The Second Fundamental Theorem of Calculus
When the upper limit of the integral is a function of x rather than x itself:
'g x
a
d f t dt f g x g xdx
g x
aA x f t dt
We can use the Second Fundamental Theorem of Calculus together with the Chain Rule to differentiate the integral:
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Example 4 (continued)Evaluate:
13
1cosxd t dt
dx
Since , , and 3cosf t t
1
23 3 1 1
1cos cosx
x x
d t dtdx
f(g(x))
1xg x 2
1'x
g x
g'(x)
3 1
2
cos x
x
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White Board ChallengeFind the derivative of the function:
2
1 secx
F x t dt
2 1 sec 2x
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White Board ChallengeIf h(t) is the rate of change of the height of a conical pile of sand in feet/hour, what does the following integral represent? Be specific.
6
2h t dt
The change in height of the pile of sand from hour 2 to hour 6.
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2006 AB Free Response 4 Form B
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1991 AB Free Response 1Let f be the function that is defined for all real numbers x and that has the following properties.i. f ''(x) = 24x – 18ii. f '(1) = –6iii. f (2) = 0
a) Find each x such that the line tangent to the graph of f at (x,f(x)) is horizontal.
b) Write the expression for f(x).c) Find the average value of f on the interval
1≤x≤3.
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2008 AB Free Response 5 Form B
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2012 AB Free Response 1