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Antiderivatives(7.4, 8.2, 10.1)
JMerrill, 2009
Review Info - Antiderivatives
General solutions:
y f(x)dx F(x) C Integrand
Variable of Integration
Constant of Integration
Review
Rewriting & Integrating – general solution
Original Rewrite Integrate Simplify
31
dxx 3x dx
2xC
2
2
1C
2x
x dx12x dx
32x
C32
322
x C3
Particular Solutions
To find a particular solution, you must have an initial condition
Ex: Find the particular solution of that satisfies the condition F(1) = 0
2
1F'(x)
x
2
1F'(x) dx
x
2x dx1x 1
C C1 x
1F(1) C
1
1F(x) C
x
0 1 C
1 C
1F(x) 1
x
Indefinite & Definite Integrals
Indefinite Integrals have the form:
Definite integrals have the form:
f (x)dxb
a
f (x)dx
7.4 The Fundamental Theorem of Calculus This theorem represents the relationship
between antiderivatives and the definite integral
Here’s How the Theorem Works
First find the antiderivative, then find the definite integral
2
3
1
4x dx4
3 44x4x dx x
4
24
1x 4 42 1 15
Properties of Definite Integrals
The chart on P. 466:
Example – Sum/Difference
Find 5
2
2
6x 3x 5 dx 5 5 5
2
2 2 2
6 x dx 3 xdx 5 dx
32 3
22
x6 x 6 2x
3
x 33 xdx 3 x
2 2
5 dx 5 x 5x
55 53 222 2
32x x 5x
2
3 3 2 232 5 2 5 2 5 5 2
2
63234 15
24352
Less Confusing Notation?
Evaluate 2
2
0
2x 3x 2 dx 23 2
0
2x 3x2x
3 2
166 4 0 0 0
3
103
Substitution - Review
Evaluate
Let u = 3x – 1; du = 3dx
43 3x 1 dx
4u du
455u
33
x 1 3dxx
C1
55C
Substitution & The Definite Integral
Evaluate
Let u = 25 – x2; du = -2xdx
5
2
0
x 25 x dx
5
2
0
5 5 12
0 0
125 x 2xdx
2
1 1udu u du
2 2
32u
C3
53
2 2
0
25 x
3
32250 1
335
32
321 u
C322
Area
Find the area bounded by the curve of f(x) = (x2 – 4), the x-axis, and the vertical lines x = 0, x = 2
2
2
0
x 4 dx23
0
x4x
3
88 0
3163
0
The answer is negative because the area is below the x-axis. Since area must be positive just take the absolute value.
163
Finding Area
Area – Last Example
Find the area between the x-axis and the graph of f(x) = x2 – 4 from x = 0 to x = 4.
2 4
2 2
0 2
x 4 dx x 4 dx 2 4
3 3
0 2
1 1x 4x x 4x
3 3
8 64 88 0 0 16 8
3 3 3
16
8.2 Volume & Average Value
We have used integrals to find the area of regions. If we rotate that region around the x-axis, the resulting figure is called a solid of revolution.
Volume of a Solid of Revolution
Volume Example
Find the volume of the solid of revolution formed by rotating about the x-axis the region bounded by y = x + 1, y = 0, x = 1, and x = 4.
Volume Example
4
2
1
V x 1 dx
43
1
x 1
3
3 3 1175 2
3 339
Volume Problem
Find the volume of the solid of revolution formed by rotating about the x-axis the area bounded by f(x) = 4 – x2 and the x-axis.
Volume Con’t
2
22
2
V 4 x dx
2
2 4
2
16 8x x dx
23 5
2
8x x16x
3 5
51264 32 64 3232 32
3 5 5 153
Average Value
Average Price
A stock analyst plots the price per share of a certain common stock as a function of time and finds that it can be approximated by the function
S(t)=25 - 5e-.01t
where t is the time (in years) since the stock was purchased. Find the average price of the stock over the first 6 years.
Avg Price - Solution
We are looking for the average over the first 6 years, so a = 0 and b = 6.
6 .01
0
125 5
6 0
te dt
The average price of the stock is about $20.15
6.01
0
1 525
6 .01
tt e
.061150 500 500
620.147
e
10.1 Differential Equations
A differential equation is one that involves an unknown function y = f(x) and a finite number of its derivatives. Solving the differential equation is used for forecasting interest rates.
A solution of an equation is a number (usually).
A solution of a differential equation is a function.
Differential Equations
Population Example
The population, P, of a flock of birds, is growing exponentially so that , where x is time in years.
Find P in terms of x if there were 20 birds in the flock initially.
0.05xdP20e
dx
Note: Notice the denominator has the same variable as the right side of the equation.
Population Cont
Take the antiderivative of each side:
This is an initial value problem. At time 0, we had 20 birds.
0.05xP 20e dx0.05xdP
20edx
0.05x 0.05x20e C 400e C
0.05
0 20 400e C
380 C
0.05xP 400e 380
One More Initial Value Problem
Find the particular solution of when y = 2, x = -1
dy2x 5
dx
dy2x 5
dx
dy2x 5dx
dx
222x
y 5x C x 5x C2
22 ( 1) 5( 1) C
6 C
2y x 5x 6
Note: Notice the denominator has the same variable as the right side of the equation.
Separation of Variables
Not all differential equations can be solved this easily.
If interest is compounded continuously then the money grows at a rate proportional to the amount of money present and would be modeled by dA
kAdt
Note: Notice the denominator does not have the same variable as the right side of the equation.
Separation of Variables
In general terms think of
This of dy/dx as the fraction dy over dx (which is totally incorrect, but it works!)
In this case, we have to separate the variables
dy f(x)dx g(y)
g(y)dy f(x)dx
G(y) F(x) C
(Get all the y’s on one side and all the x’s on the other)
Example
Find the general solution of
Multiply both sides by dx to get
2dyy x
dx
2y dy x dx
2y dy x dx 2 3y x
C2 3
2 32y x 2C
3
Lab 4 – Due Next Time on Exam Day
1. #34, P471 9. #25, P523 2. #59, P440 10. #35, P523 3. #22, P471 11. #3, P629 4. #45, P439 12. #7, P629 5. #11, P439 13. #19, P630 6. #13, P471 14. #27, P630 7. #27, P439 15. #43, P472 8. #17, P522