Post on 19-Jan-2016
Day 1
Section 4.4
Antidifferentiation → Indefinite Integral:
(the family of functions)
Definite Integration → Definite Integral:
a number
On the calculator: Alpha – Window − 4
'f x dx f x C
'b
af x dx
If a function f is continuous on the closed interval [a, b], then
If f (x) = F '(x) or F(x) is the antiderivative of f (x), then FTOC can be written as
Fundamental Theorem of Calculus
'b
af x dx f b f a
b
af x dx F b F a
An Illustration of FTOC
2Let ' 2 and .f x x f x x 2
1Find 2 .x dx
1 2
1
2
3
4
2
12x dx 1
1 2 42
3 Now find 2 1 .f f
2 22 1 2 1f f 3
2
1' 2 1f x dx f f
Examples
8 32
0
3 32 2
821. 1 1
03
2 21 8 1 0
3 32 2 2 1
27 1 18 173 3 3 3
x dx x
2. 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.
1 2
-1
1
2
3
4 2 2
02 3 2x x dx
2 2
02 3 2x x dx
23 2
0
2 32
3 2
x xx
3 2
2 2 3 22 2 0
3 2
16
6 43
16 6 10
3 3 3
3 3
44
3. cos sin
sin sin3 4
3 2
2 2
x dx x
When using u-substitution with a definite integral, it is often convenient to determine the limits of integration for the variable u rather than to convert the antiderivative back to the variable x and evaluate at the original limits.
Day 2
If the function u = g(x) has a continuous derivative on the closed interval [a, b] and f is continuous on the range of g, then
Change of Variables for Definite Integrals
'
b g b
a g af g x g x dx f u du
The new upper and lower limits of integration:
When x = 0, then u = 02 + 1 = 1. (lower limit)
When x = 1, then u = 12 + 1 = 2. (upper limit)
Example 1
31 2
0Evaluate: 1x x dx
2 1Let + 1 so 2 and .
2 u x du x dx du xdx
32
1
1Evaluate:
2 u du
24
1
1 1
2 4
u
24 4
1
12 1
8
15
8
Example 2
2
1Evaluate: 1 2 x x dx
Let 2 . u x
2 u x
du dx2 x u
1 2 1 1 x u u
The new upper and lower limits of integration:
When 1, then 2 1 1 x u
When 2, then 2 2 0 x u
2 0
1 11 2 1 x x dx u u du
0
1 1 u u du
1 1
2
0 1 u u du
2 20
3 5
4
15
1
1 3 3 512 2 2 2
00
2 2
3 5 u u du u u
2
1Evaluate: 1 2 x x dx
Day 3
Given
with the initial condition y(2) = -1. Find y(3).
Method 1
Integrate use the initial
condition to find C. Then write the particular
solution , and use your particular solution to find y(3).
23 4 5 dy
x xdx
23 4 5 , y x x dx
23 4 5 y x x dx
3 23 45
3 2
x xx C
3 22 5 x x x C
232 2 2 2 5 2 y C
1 8 8 10 C7 C
233 3 2 3 5 3 7 y
3 22 5 7 y x x x
23
Method 2: Use the Fundamental Theorem of Calculus
' b
af x dx f b f a
3 2
23 4 5 3 2 x x dx f f
3 2
23 2 3 4 5 f f x x dx
33 2
21 2 5 x x x
1 27 18 15 8 8 10 1 30 6 23
Sometimes there is no antiderivative so we have to use Method 2 and our graphing calculator.
3. f ′(x) = sin(x2) and f (2) = -5. Find f (1).
2 2
1sin 2 1 x dx f f
2 2
1sin 5 1 x dx f
2 2
11 5 sin f x dx
5.495
Graph of f ′
4. The graph of f ′ consists of two line segments and a semicircle as shown on the right. Given that f (-2) = 5, find:
(a) f (0)
(b) f (2)
(c) f (6)
Graph of f ′
(a) f (0)
0
2' 0 2
f x dx f f
0
20 ' 2
f f x dx f
0 4 5 9 f
Graph of f ′
(b) f (2)
2
2' 2 2
f x dx f f
2 8 5 13 f
2
22 ' 2
f f x dx f
Graph of f ′
(c) f (6)
6
2' 6 2
f x dx f f
6 8 2 5 13 2 f
6
26 ' 2
f f x dx f
5. The graph of f ′ is shown. Use the figure and the fact that f (3) = 5 to find:
(a) f (0)
(b) f (7)
(c) f (9)
(a) f (0)
3
0' 3 0 f x dx f f
0 5 4 1 f
3
00 3 ' f f f x dx
(b) f (7)
7
3' 7 3 f x dx f f
7 9 5 4 f
7
37 ' 3 f f x dx f
(c) f (9)
9
7' 9 7 f x dx f f
9 2 4 2 f
9
79 ' 7 f f x dx f
Then sketch the graph of f using the points from a through c.
+ area → f is increasing
− area → f is decreasing
Relative maximum at (3, 5)
Relative minimum at (7, -4)
0 1f
9 2f
6. A pizza with a temperature of 95°C is put into a 25°C room when t = 0. The pizza’s temperature is decreasing at a rate of r(t) = 6e-0.1t °C per minute. Estimate the pizza’s temperature when t = 5 minutes.
5 0.1
06 5 0 te dt t t
5 95 t
5 0.1
05 6 95 tt e dt
23.608 95 71.392 C
Day 4
Applications of the First Fundamental Theorem
If f is continuous on [a, b] and differentiable on (a , b) such that:
Mean Value Theorem for Derivatives
a c1 c2 b
'
f b f af c
b a
Mean Value Theorem for Derivatives
In Section 4.2, we saw that the area of a region under a curve is greater than the area of an inscribed rectangle and less than the area of a circumscribed rectangle.
The Mean Value Theorem for Integrals states that somewhere between the inscribed and circumscribed rectangles there is a rectangle whose area is precisely equal to the area of the region under the curve
b
Too small
aa b
c
b
af x dx f c b a
L W
Mean Value Theorem for Integrals
Too big
ba
Exact Area
ba
Area of Rectangle
If f is continuous on [a, b] and differentiable on (a , b) such that:
Mean Value Theorem for Integrals
b
af x dx f c b a
Example 1
3
Find the value(s) of guaranteed by the Mean Value
9Theorem for Integrals for on the interval
1, 3 .
c
f xx
3
31
93 1dx f c
x
b
af x dx f c b a
3 3
19 3 1x dx f c
3
21
92
2f c
x
2 2
9 92
2 3 2 1f c
4 2 f c
2f c
3
92
c
3 9
2c
39
2c
The value f (c) given in the Mean Value Theorem for Integrals is called the average value of f on [a, b].
1 b
af c f x dx
b a
b
af x dx f c b a
Average Value of f
Find the average value of f (x) = 3x2 – 2x on the interval [1, 4].
Example 2
4 2
1
13 2
4 1f c x x dx
1 b
af c f x dx
b a
43 2
1
1
3x x
14 1
3F F
164 16 1 1
3
148 16
3
The First Fundamental Theorem of Calculus could also be called the “Total Change Theorem”. If you are given the rate of change of a function, f ′(x), you can find the accumulated change in the function f (x).
First Fundamental Theorem of Calculus
'b
af x dx f b f a
or
'b
af b f a f x dx
Applications Examples
1. Suppose that C(t) represents the cost per day to heat your house measured in dollars per day, where t is measured in days and t = 0 corresponds to January 2, 1993. Interpret
90 90
0 0
1 and .
90 0C t dt C t dt
The units for the integral are
(dollars/day)(days) = dollars.
The integral represents the total cost in dollars to heat your house for the first 90 days of 1993 (January through March).
The second expression is the average cost per day to heat your house for the first 90 days of 1993.
The units would be in dollars/day, the same units as C(t).
90
0C t dt
2. If V(t) is the volume of water in a reservoir at time t, then its derivative V ′(t) is
the rate at which water flows into the reservoir at time t.
is the change in the amount of water in the reservoir between time t1 and t2.
2
12 1So '
t
tV t dt V t V t
3. If the rate of growth of a population is dn/dt,
is the increase in population during the time period from t1 to t2.
2
12 1then
t
t
dn dt n t n tdt
4. If w′(t) is the rate of growth of a child in pounds per year, what does
It represents the increase in weight of a child from 5 years old to 10 years old.
The units are pounds.
10
5' represent?w t dt
10
5' 10 5w t dt w w
5. If oil leaks from a tank at a rate of r(t) gallons per minute at time t, what does
It represents the decrease of oil in the tank over the 1st 2 hours or 1st 120 minutes. The units are gallons.
120
0 represent?r t dt
120
0120 0r t dt R R
6. A honey bee population starts with 100 bees and increases at a rate n′(t) bees per week.
represents the total amount of bees after 15 weeks.
15
0What does 100 ' represent?n t dt
15
0' 15 0n t dt n n
15
0' 15 100n t dt n
15
015 ' 100n n t dt
7. A cup of coffee at 90° is put into a 20° room when t = 0. The coffee’s temperature is changing at a rate of r(t) = -7e-0.3t °C per minute with t in minutes. Find the coffee’s temperature when t = 10.
10 0.3
07 te dt 10 0R R
10 90R
10 0.3
010 90 7 tR e dt
90 22.172 67.828
8. The rate at which water is being pumped into a tank is r(t) = 20e0.02t where t is in minutes and r(t) is in gallons per minute.
a. Approximately how many gallons of water have been pumped into the tank in the first five minutes?
5 0.02
0
5 0.02
0
5 0.02
0
20 5 0
20 5 0
5 20
5 105.171 gallons
t
t
t
e dt R R
e dt R
R e dt
R
5 0.02
0
120
5 0te dt
105.171
5
b. Find the average rate at which the water is being pumped into the tank during the first five minutes.
21.034 gallons/minute
9. A faucet was turned on at t = 0, and t minutes later water was flowing into a barrel at a rate of t2 + 4t gallons per minute, 0 ≤ t ≤ 5.
a. Use a calculator to find how much water was added to the barrel during these 5 minutes?
5 2
04 91.667 gallonst t dt
5 2
0
1 91.6674
5 0 518.333 gallons/minute
t t dt
b. Find the average flow rate for these five minutes.
Day 6
Functions Defined by Integrals
The Definite Integral as a Number
b
af x dx F b F a
Constant
Constant f is a functionof x
The Definite Integral as a Function of x
x
aF x f t dt
F is a function of x
Constant f is a functionof t
1. A function g is defined on the interval [0, 6] by
where f is the function graphed on the next slide. Fill in the following table of values.
0
xg x f t dt
x g(x)
0
1
2
3
4
5
6
0
xf t dt
0
0f t dt 0
1
0f t dt ½
2
0f t dt 1
3
0f t dt ½
4
0f t dt 0
5
0f t dt ½
6
0f t dt 1
0
xg x f t dt
Graph of fGraph of g
2. Let
where f is the function graphed on the next slide. Fill in the following table of values.
0
xg x f t dt
x g(x)
-2
0
2
4
5
0
xf t dt
2
0f t dt
-
0
0f t dt 0
2
0f t dt
4
0f t dt − 1
5
0f t dt − ½
0
2f t dt
Graph of f
3. The graph of a function f consists of a quarter circle and line segments. Let g be the function given by
0
.x
g x f t dt 0g
0
00f t dt
1g 1
0f t dt
2g
0
11f t dt
2
0f t dt 5g
5
04f t dt
Graph of f
b. Find all values of x on the open interval (-1, 5) at which g has a relative maximum. Justify your answer.
g′ = f = 0 or undef.
f = 0 at x = 2
g has a relative max.
at x = 2 because g′ = f
which changes from
positive to negative at
x = 2.
Graph of f
c. Find the absolute minimum value of g on [-1, 5] and the value of x at which it occurs. Justify your answer.
The candidates for the absolute minimum are at the endpoints and the critical numbers.
g(-1) = -1
g(2) = g(5) = − 4
The absolute minimum is at x = -1.
Day 7
4. Use the Fundamental Theorem of Calculus to rewrite the following accumulation functions. Then find the derivative of each function.
2
1a.
xf x t dt
3
13
xt
1F x F 3 1
3 3
x
2'f x x
6
b. cosx
g x t dt
6
sinx
t
6G x G
sin sin6
x
' cosg x x
1sin
2x
4 2c.x
h x t dt 43
3x
t
4H H x 3 34
3 3
x
2'h x x
364
3 3
x
The Second Fundamental Theorem of Calculus
x
a
df t dt
dx f x
a
x
df t dt
dx f x
22
15.
xt dt
23
13
xt
2 1F x F
321
3 3
x
6 1
3 3
x
22
1
xdt dt
dx 52x
Second Fundamental Theorem of Calculus (Chain Rule Version)
u x
a
df t dt
dx f u du
26.
x
x
dt dt
dx
0 2
0
x
x
dt dt t dt
dx
2
0 0
x xdt dt t dt
dx
2 2f x f x
2 2x x
7. Use the Second Fundamental Theorem to evaluate:
2
3a. 1
xdt dt
dx 21 x
3
2b. tan
xdt dt
dx 3tan x
3
1
1c.
1
xddt
dx t
23
13
1x
x
2
3
3
1
x
x
sin 3 2
2d. 1
xdt dt
dx 3 21 sin cosx x
d. Find the x-coordinate of each point of inflection on the graph of g in (-1, 5). Justify your answer.
x = 0 and x = 3 are
points of inflection
because g′ changes from
increasing to decreasing (x = 0) and decreasing to increasing (x = 3).
0
x
g x f t dt
' g x f x