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HL Math 1 – Calculus - Santowski Lesson 45 - Antiderivatives and the Rules of Integration 1/6/2016...
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Transcript of HL Math 1 – Calculus - Santowski Lesson 45 - Antiderivatives and the Rules of Integration 1/6/2016...
HL Math 1 – Calculus - Santowski
Lesson 45 - Antiderivatives and the Rules of Integration
04/21/23
1
Calculus - Santowski
A function F is an antiderivative of f on an interval I if F’(x)=f(x) for all x in I.
Let’s use an example to figure out what this statement means…
Suppose we know f(x) = 2x and we want to find its antiderivative, F.
If f(x) = 2x, then F’(x) = 2x. So we know the derivative of F.
Think backwards, what function has a derivative equal to 2x ?
F(x) = x² !!!
To find the antiderivative, do the reverse of finding the derivative.
Is F(x) = x² the only function whose derivative is 2x ? Or in otherwords, is F(x) = x² the only antiderivative of 2x ? 04/21/232Calculus - Santowski
Theorem 1: Let G be an antiderivative of a function f. Then, every antiderivative F of f must be of the form F(x) = G(x) + C, where C is a constant.
Example 1: Let F(x) = x² + 4 and let G(x) = x²- 1 Then F’(x) = 2x and G’(x)= 2x
Thus both F and G are antiderivatives of f(x) = 2x.Note two functions which have the same derivative will only differ by the constant.
This means the antiderivative of a function is a family of functions as pictured on the next slide.
So the answer to the question is F(x) = x²the only antiderivative of2x is NO!!
04/21/233Calculus - Santowski
x
y
F(x)= x2 - 3
F(x)= x2 - 2
F(x)= x2 - 1
F(x)= x2
F(x)= x2 +1
When you integrate (which is another name for finding the antiderivative) you get a family of curves. Each of these is an antiderivative of f(x)=2x.
04/21/234Calculus - Santowski
The Indefinite Integral
The process of finding all antiderivatives of a function is called antidifferentiation, or integration.
is the symbol used for integration and is called the integral symbol.
We write
€
f(x)dx=F(x)+C
This is called an indefinite integral, f(x) is called the integrand and C is called the constant of integration.
dx is the differential of x which denotes the variable of integration
04/21/235Calculus - Santowski
Basic Integration Rules
€
kdx=kx+CRule 1: (k, a constant)
Example 2:
€
2dx=2x+C
Example 3:
€
π2dx=π2x+C
Keep in mind that integration is the reverse of differentiation. What function has a derivative k?
kx + C, where C is any constant.
Another way to check the rule is to differentiate the result and see if it matches the integrand. Let’s practice.
04/21/236Calculus - Santowski
Before we list Rule 2, let’s go back and think about derivatives.
When we used the power rule to take the derivative of a power, we multiplied by the power and subtracted one from the exponent.
Example:
€
ddx
x3( ) =3x2
Since the opposite of multiplying is dividing and the opposite of subtracting is adding, to integrate we’d do the opposite. So, let’s try adding 1 to the exponent and dividing by the new exponent.
Check by differentiating the result:
€
ddx
14
x4 ⎛
⎝ ⎜
⎞
⎠ ⎟=44
x3 =x3
Since we get the integrand we know it works.
€
x3 dx=x3+1
3+1=14
x4Integrating:
04/21/237Calculus - Santowski
Rule 2: The Power Rule C1n
xx
1nn
n 1
Example 4: Find the indefinite integral
€
t3 dt
Solution:
€
t3 dt=t4
4+C
Example 5: Find the indefinite integral
€
x32 dx
€
x32 dx=
x32+1
32+1
+C=X
52
52
+C=25
X52 +CSolution:
Basic Integration Rules
04/21/238Calculus - Santowski
Example 6: Find the indefinite integral dxx13
Solution:
€
1x3 dx= x−3 =
x−3+1
−3+1+C=
x−2
−2+C=
−12x2
+C
Example 7: Find the indefinite integral dx1
€
1dx=x+CSolution:
Example 8: Find the indefinite integral dxx3 2
Solution:
€
−3x−2 dx=−3 x−2 =−3x−2+1
−2+1+C=
−3x−1
−1+C=
3x+C
Here are more examples of Rule 1 and Rule 2.
04/21/239Calculus - Santowski
Rule 3: The Indefinite Integral of a Constant Multiple of a Function
€
( )cf x dx=c ( ) ,f x dxcisaconstant
Rule 4: The Sum Rule (or difference)
€
f x( )+g x( )[ ] dx= ( )f x dx+ ( )g x dxf x( )−g x( )[ ] dx= ( )f x dx− ( )g x dx
Rule 5:
€
ex dx=ex+C
Rule 6:
€
1x
dx=lnx+C
Basic Integration Rules
To check these 2 rules, differentiatethe result and you’ll see that it matchesthe integrand. 04/21/2310Calculus - Santowski
Integrate and check your answer by taking the derivative.
€
( ) a 3+x4 +exdx
€
( ) b (2x−1x
+3x2
+ x+3ex)dx
€
( ) c 2x−4( )1x2
+3x
⎛
⎝ ⎜
⎞
⎠ ⎟dx
04/21/2311Calculus - Santowski
Here is the solution for Q(a)in detail.
€
( ) a 3+x4 +exdx = 3dx + x4dx + exdx
= 3x +x4+1
4+1+ ex +C
=3x+15
x5 +ex +C
04/21/2312Calculus - Santowski
Q(b) Integrate
€
(2x−1x
+3x2
+ x+3ex)dx
Using the sum rule we separate this into 5 problems.
€
2xdx−1x
dx+3x2 dx+ x dx+ 3 exdx
€
2xdx−1x
dx+3x2 dx+ x dx+ 3 exdx=x2 −lnx−
3x+23
x32 +3ex+C
You may be wondering why we didn’t use the C before now. Let’s say that we had five constants . Now we add all of them together and call them C. In essence that’s what’s going on above.
54321 CCCCC
ANSWER:
04/21/2313Calculus - Santowski
(c) Integrate
€
2x−4( )1x2
+3x
⎛
⎝ ⎜
⎞
⎠ ⎟dx
€
Multiplythetwofactorsintheradicandtogether
andcombinethelike .terms
2x+6−
4x2
−12x
dx= 6−10x−4x2
dx
€
NextuseRule4 tointegrateeachterm .separately
6dx−101xdx−4 x−2dx
6x−10lnx−4x−1
−1+C
6x−10lnx+4x+C
04/21/2314Calculus - Santowski
Differential Equations
A differential equation is one which has a derivative expression in it.
For example: f’(x)=2x+1 is a differential equation.When we integrate this type of equation we get the general solution which contains the constant, C .
To find a particular solution we need another piece of information. It could be a point that we know the function passes through. We call this piece of information the initial condition.
A typical problem might be:f’(x)= 3x2 - 4x + 8f(1)= 9
04/21/2315Calculus - Santowski
Let’s look at the solution to the problem:f’(x)= 3x2 - 4x + 8f(1)= 9
Solution: First integrate both sides:
Cx82x4
3x3
xf23
)(
Simplify: Cx8x2xxf 23 )(
Now find C by using the initial condition.Substitute 1 for x and 9 for f(x)
C2
C79
C8219
C181219 23
This gives the particular solution.
2x8x2xxf 23 )(04/21/2316Calculus - Santowski
Review - Basic Integration Rules
Rule 1: (k, a constant)
€
kdx=kx+CRule 2: The Power Rule C
1nx
x1n
n
Rule 3: The Indefinite Integral of a Constant Multiple of a Function
€
( )cf x dx=c ( ) ,f x dxcisaconstantRule 4: The Sum Rule (or difference)
€
f x( )+g x( )[ ] dx= ( )f x dx+ ( )g x dxf x( )−g x( )[ ] dx= ( )f x dx− ( )g x dx
Rule 5:
€
ex dx=ex+C
Rule 6:
€
1x
dx=lnx+C04/21/2317Calculus - Santowski
Integration Formulas for Trigonometric Functions
Cxdxxx
Cxdxxx
Cxdxx
Cxdxx
Cxdxx
Cxdxx
csccotcsc
sectansec
cotcsc
tansec
sincos
cossin
2
2
04/21/23
18
Calculus - Santowski
Examples.
€
( ) a (3cscxcotx−7sec2 x)dx
€
( ) b3 tanθ −4cos2 θ
cosθdθ
04/21/23
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Calculus - Santowski
Solution to Example (a).
€
(3cscxcotx−7sec2 x)dx=3 cscxcotxdx−7 sec2 x dx=3 −cscx+C1[ ] −7 tanx+C2[ ]
=−3cscx−7 tanx+C
04/21/23
20
Calculus - Santowski
Solution to Example (b)
€
3 tanθ −4cos2 θcosθ
dθ
=31
cosθtanθdθ −4 cosθdθ
=3 secθ tanθdθ −4 cosθdθ=3secθ −4sinθ +C
04/21/23
21
Calculus - Santowski
Exercises:
θθθ d
dxx
x
dyy
yy
dxxx
dyyy
)tan3cot2(.5
cos
sin.4
12.3
)1(.2
)32(.1
22
2
24
23
04/21/23
22
Calculus - Santowski
Applications of Indefinite Integrals
1. Graphing
Given the sketch of the graph of the function, together with some function values, we can sketch the graph of its antiderivative as long as the antiderivative is continuous.
04/21/23
23
Calculus - Santowski
Example 1. Given the sketch of the function f =F’(x) below, sketch the possible graph of F if it is continuous, F(-1) = 0 and
F(-3) = 4.
-1
4
5
-3 -2 -1 0 1 2 3 4 5
1
-4-5
3
2
-2
-3
-4
-5
F(x) F’(x)
F’’(x)
Conclusion
X<-3 + - Increasing, Concave down
X=-3 4 0 - Relative maximum
-3<x<-2
- - Decreasing, Concave down
X=-2 - 0 Decreasing, Point of inflection
-2<x<-1
- + DecreasingConcave up
X=-1 0 0 + Relative minimum
X>-1 + + Increasing,Concave up
04/21/23
24
Calculus - Santowski
The graph of F(x)
-1
4
5
-3 -2 -1 0 1 2 3 4 5
1
-4-5
3
2
-2
-3
-4
-5
04/21/23
25
Calculus - Santowski
1. Boundary/Initial Valued Problems
There are many applications of indefinite integrals in different fields such as physics, business, economics, biology, etc.
These applications usually desire to find particular antiderivatives that satisfies certain conditions called initial or boundary conditions, depending on whether they occur on one or more than one point.
Applications of Indefinite Integrals
04/21/23
26
Calculus - Santowski
Example 2.
6 1dy
xdx
Suppose we wish to find a particular antiderivative satisfying the equation
and the initial condition y = 7 when x = 2.
04/21/23
27
Calculus - Santowski
Sol’n of Example 2
2
2
(6 1)
(6 1)
3
2 7,
7 3(2) 3 7
dy x dx
dy x dx
y x x C
but x when y then
C C
23 7y x x Thus the particular antiderivative desired,
04/21/23
28
Calculus - Santowski
Example 3.
The volume of water in a tank is V cubic meters when the depth of water is h meters. The rate of change of V with respect to h is π(4h2 +12h + 9), find the volume of water in the tank when the depth is 3m.
04/21/23
29
Calculus - Santowski
Sol’n of Example 3
2
2
32
32
32
32 3
4h 12h 9
4h 12h 9
46h 9h
3
4(0 )0 6(0 ) 9(0)
3
0
46h 9h
3
4(3 )6(3 ) 9(3) 207
3
dV
dh
dV dh
hV C
C
C
hThus V
V m
π
π
π
π
π
π π
Volume V=0 if depth h =0
04/21/23
30
Calculus - Santowski
The Differential Equations
Equation containing a function and its derivative or just its derivative is called differential equations.
Applications occur in many diverse fields such as physics, chemistry, biology, psychology, sociology, business, economics etc.
The order of a differential equation is the order of the derivative of highest order that appears in the equation.
The function f defined by y= f(x) is a solution of a differential equation if y and its derivatives satisfy the equation.
04/21/23
31
Calculus - Santowski
If each side of the differential equations involves only one variable or can be reduced in this form, then, we say that these are separable differential equations.
Complete solution (or general solution)y = F(x) + C
Particular solution – an initial condition is given
The Differential Equations
04/21/23
32
Calculus - Santowski
Example 4.
342
2
xdx
yd
Find the complete solution of the differential equation
04/21/23
33
Calculus - Santowski
Solution to Example 4.
2
2
21
4 3
(4 3)
(4 3)
2 3
d y dyx
dx dxdy x dx
dy x dx
y x x C
21
21
3 2321 23 2
2 3
(2 3 )
dyx x C
dx
dy x x C dx
y x x C x C
2
2
d y dyddx dxdx
dydxlet y
342
2
xdx
ydFind the complete solution of the differential equation
04/21/23 34Calculus - Santowski
Example 5.
Find the particular solution of the differential equation below for which y=2 and y’=-3 when x=1.
342
2
xdx
yd
04/21/23
35
Calculus - Santowski
Solution to Example 5.
21
21
1
2 3
3 2(1) 3
8
y x x C
x C
C
3 2321 23 2
3 23223 2
472 6
2 (1) (1) 8(1)
y x x C x C
C
C
3 23 4723 2 68y x x x
Find the particular solution of the differential equation in Ex. 14 for which y=2 and y’=-3 when x=1.
04/21/23 36Calculus - Santowski
Example 6.
A stone is thrown vertically upward from the ground with an initial velocity of 20ft/sec.
(a) How long will the ball be going up?
(b) How high will the ball go?
(c) With what velocity will the ball strike the ground?
04/21/23
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Calculus - Santowski
ANS to Example 6.
A stone is thrown vertically upward from the ground with an initial velocity of 20ft/sec.
(a) How long will the ball be going up?Ans. 0.625 sec
(b) How high will the ball go?Ans. 6.25 ft
(c) With what velocity will the ball strike the ground?
Ans. 20 ft/sec
04/21/23
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Calculus - Santowski
Other Properties of Indefinite Integrals
04/21/23Calculus - Santowski
39
And two other interesting “properties” need to be highlighted:
Interpret what the following 2 statement mean:
Let f(x) = x3 - 2x
What is the answer for f `(x)dx ….?
What is the answer for d/dx f(x)dx ….. ?
Examples with Motion
04/21/23Calculus - Santowski
40
An object moves along a co-ordinate line with a velocity v(t) = 2 - 3t + t2 meters/sec. Its initial position is 2 m to the right of the origin.
(a) Determine the position of the object 4 seconds later
(b) Determine the total distance traveled in the first 4 seconds
Examples – “B” Levels
04/21/23Calculus - Santowski
41
Sometimes, the product rule for differentiation can be used to find an antiderivative that is not obvious by inspection
So, by differentiating y = xlnx, find an antiderivative for y = lnx
Repeat for y = xex and y = xsinx
Integration by Chain Rule/Substitution
For integrable functions f and g
where is an F antiderivative of f and C is an arbitrary constant.
( ( ))[ '( ) ] ( ( ))f g x g x dx F g x C
04/21/23
42
Calculus - Santowski
Example 1.
14
54
54
2 34
3 24
3 2
3
54
385
36 6 5
2 6 5 (18 )
2 (6 5) (18 )
(6 5)2
(6 5)
x x dx
x x dx
x x dx
xC
x C
Let g(x) = 6x3+5
g’(x)=18x2
04/21/23
43
Calculus - Santowski
Example 1. Take 2!
14
54
54
54
2 3 3 24 4
4
54
85
385
36 6 5 2 6 5 (18 )
2
2 ( )
2
(6 5)
x x dx x x dx
u du
u du
uC
u C
x C
Let u = 6x3 + 5
du = 18x2 dx
04/21/23
44
Calculus - Santowski
Example 2.
3 3
7 74 4
74 3
64
64
2 1 1 2(2 1)
22 2
12 2(2 1)
2
21
2 61
12 2
t t dxdx
t t t t
t t t dx
t tC
Ct t
Let g(t) = t4 + 2t
g’(t) = 4t3 + 2
= 2(2t3 + 1)
04/21/23
45
Calculus - Santowski
Example 3.
5 2 12
22 2 12
2 12
14 13 12
15 14 131 2 115 14 13
2 15 2 14 2 131 1 115 7 13
( 1) 2
( 1) 2
( 1)
( 2 )
( 1) ( 1) ( 1)
x x dx
x x xdx
u u du
u u u du
u u u C
x x x C
Let u = x2 -1
du = 2x dx
x2 = u+1
04/21/23
46
Calculus - Santowski
Example 4.
Let u = 2 – cos2x
du = 0 – (-sin2x)(2dx)
=2sin2xdx
Cx
Cu
duu
duu
xdxx
dxxx
2/331
2/332
21
2/121
21
21
)2cos2(
)(
)(
)2sin2(2cos2
2cos22sin
04/21/23
47
Calculus - Santowski
Example 5.
xdxx
dxxx
dxxx
xx
dxx
x
x
x
dxxx
2csc2sec
2sin2cos
1
2sin2cos
2cos2sin
2sin
2cos
2cos
2sin
)2cot2(tan
22
22
222
2
2
04/21/23
48
Calculus - Santowski
Example 6
Cxx
Cxx
xdxdxxx
xdxdxxx
xdxxdxx
dxxx
xdxx
2tan2cot
2tan)2(tan
2sec22sec2)2(tan
2sec2sec)2(tan
2sec2cot2sec
)12(cot2sec
2csc2sec
21
21
21
21
22122
21
222
222
22
22
1
04/21/23
49
Calculus - Santowski
Exercises:
drr
r
xx
dxxx
7
3 22
62
)1(
2.3
)49(5.2
)12(7.1
424
2
22
)123(
)13(.6
3sin21
3cos.5
3cot3csc.4
xx
dxxx
dxx
x
dyyyy
04/21/23
50
Calculus - Santowski