**Unformatted text preview: **1. Techniques of Integration (Chapter 7) Integration by Parts Trigonometric Integrals Trigonometric Substitution Integration of Rational Functions Definite vs. Indefinite Integrals Strategy for Integration Improper Integrals Convergence vs Divergence Comparison Theorem
2. Applications of Integration (Chapter 8) Arc Length Area of a Surface Revolution
3. Differential Equations (Chapter 9) Modelling with DE Direction Fields Separable Equations Linear Equations Applications e.g. population growth
4. Parametric Equations & Polar Coordinates (Chapter 10) Sketching curves defined by parametric equations Calculus with parametric equations Polar Coordinates
5. Sequences & Series (Chapter 11) Sequences Divergence Test The Integral Test and Estimates of Sums The Comparison Tests Alternating Series Absolute Convergence Ratio Test Root Test Power Series Representations of Functions as Power Series Taylor and Maclaurin Series Taylor’s Inequality Chapter 7: Techniques of Integration
I. Integration by Parts = −
Example: 2 Let = 2 , = 2 = , = 2 = 2 − 2 = 2 − 2 () Evaluate this integral using integration by parts again
Let = , = = , = − = − − − = − + = − + + Substitute back into the original equation: 2 = 2 − 2(− + + )
II. Trigonometric Integrals
Strategy for evaluating , ≥ 0 a) If m is even and n is odd:
sin 2+1 = sin = (cos 2 ) 1 − 2 Let = , = = 1 − 2 Example: 2 −1 Expand, simplify, integrate cos 3 = sin2 1 − 2 = 2 1 − 2 2 2 = 6 − 24 + 2 7
5
3
=
−2 +
+ 7
5
3
Substitute = 7 5 3 −2
+
+ 7
5
3 = b) If m is odd and n is even: Same as a) but we want everything as a function of cos instead of sin.
Example: 3 cos 2 = sin2 (cos 2 ) (1 − cos2 ) (cos 2 ) = Let = , = − Substitute, expand, simplify, integrate.
c) If m and n are both odd, use either a) or b).
d) If m and n are both even, use trig identities!
Example: 4 = (sin2 )2 1 − 2 2 2
1
=
(1 − 2 cos 2 + cos 2 2) 4
= 1 Substitute the trig identity cos 2 2 = 2 (1 + 4)
= 1
4 1 − 22 + 1
1 + 4
2 1
3
1
− 22 + 4 4
2
2
1 3
1
= ( − sin 2 + 4) + 4 2
8
= We can use similar strategies for evaluating tan sec using the identity 2 = 1 + tan2 a) If n is even: tan sec 2 =
= tan ( 2 ) −1 sec 2 tan (1 + tan2 ) −1 sec 2 Let = , = sec 2 = (1 + 2 ) −1 expand, simplify, integrate Example: tan6 4 = tan6 sec 2 sec 2 = Let = , = sec 2 tan6 (1 + tan2 ) sec 2 = 6 1 + 2 7
9
+
+ 7
9
tan7 tan9 =
+
+ 7
9
= b) If m is odd: tan2+1 sec =
= (tan2 ) sec −1 (sec 2 − 1) −1 Let = sec , = = (2 − 1) −1 Expand, simplify, integrate III. Trig Substitution
Expression Substitute
2 − 2
2 + 2 = 2 − 2
Example: = = 2 − 2 Let = sec = 2 sec 2 − 2 = 2 (sec 2 − 1) =
2 tan2 = tan With this in mind, let’s evaluate the integral 2 − 2 Let = , = = sec tan tan = = ln + + = ln | + 2 − 2
|+ IV. Integration of Rational Functions = () f(x) is proper if degree of P < degree of Q, improper otherwise. () 1. Make f(x) proper if f(x) is improper. ()
()
= +
()
() =
Example: = 3 3 −12 2 −38 degree of P = 3, degree of Q = 1, must make it improper. −3 Factor P(x) by Q(x) – remember factoring polynomials by long vision? It’s back to haunt you. After
factoring, f(x) becomes = 3 2 − 3 − 9 − 65
−3 2. Factor Q(x) as far as possible.
Example: = 4 − 16
= 2 − 4
3. Express R(x)/Q(x) as a sum of + 2 + 4 = − 2 +2 2 + 4 + or ( 2 + +) Four cases to consider:
a) Q(x) is factored and no factor is repeated
b) There are repeats
c) Q(x) has irreducible factor 2 + + d) Q(x) contains a repeated irreducible factor
V. Strategies for Integration
When given an integration problem, look for:
Square roots use trigonometric substitution
Obvious Substitutions U substitution
Trig functions trig integrals
Rational functions Use integration for rational functions
There’s going to be instances where you may start a problem with one technique and realize that will not
work. That’s okay! Try again with another technique. PRACTICE PRACTICEPRACTICE. Become the master
at integration!
VI. Improper Integrals
Allows us to estimate the area under a curve by calculating a definite integral where the lower or upper
boundary is ±∞.
Example: ∞ 1
1 2 lim → ∞ 1
1 2 Improper integral because upper boundary = ∞. 1
, 1
→∞ 1
= lim →∞ 1 − = 1 Limit exists, therefore it is convergent.
= lim − Example: ∞1
1 = lim →∞ 1
1 = lim ln
()
→ ∞ = lim ln − 1
()
→ ∞ = lim →∞ ln = ∞Limit is divergent.
Comparison Theorem: For () ≥ () ≥ 0, if f(x) is convergent, then g(x) is also convergent. If g(x) is
divergent, f(x) is also divergent. This theorem allows us to evaluate improper integrals where the integral
may be very difficult to evaluate.
Example: ∞ 1+ −
1 1+ − and we know from above that > 1 ∞1
1 is divergent -- therefore ∞ 1+ −
is
1 also divergent. Chapter 8: Applications of Integration
I. Arc Length 1 + ( ′ )2 () =
3 Example: What is the arc length of = 2 from1 to 4?
4 3 1
1+
2
2 =
1
1 2 9
1 + ( ) 4 =
0
1 Evaluate the integral using u substitution. = 27 (80 10 − 13 13)
II. Area of a Surface Revolution
= 2 , where ds = Arc Length Example: Find the surface area obtained by rotating =
= about the x-axis from 4 to 9. 2 9 = 2 1+
4
9 = 2 4 1+ 1
2 2 1 4 We want to simplify this expression – get common denominator 9 = 2 4 + 1 4 4
9 = 2 4 + 1 2 4
9 = 4 + 1 4
3 Evaluate using u substitution. 3 = 6 (37 2 − 17 2 ) Chapter 9: Differential Equations
I. Modelling with DE
Show that = − cos − is a solution to ′ = + 2 .
Substitute this and y into ′ = + 2 . ′ = − + − 1 − + − 1 = − − + 2 − + 2 − = − + 2 − Therefore, y is a solution to the equation.
III. Separable Equations Solve the differential equation = 2
2 if y(0) = 2. Collect all y terms on one side, all x terms on the other side. Integrate both sides, then solve for y. 2 = 2 2 = 2 3
3
+ 1 =
+ 2
3
3
3 = 3 + 3 0 =
=8
∴ = 3 0+ = 2
3 + 8 IV. Linear Equations
Look for the form ′ + = ()
Method: ′ + = , where = ′ + = ′ ( ) Integrate both sides, solve for y.
Example: ′ − = . Here, () = −1, () = , () = −1 = − ′ = ′ + ′ − ′ = − = − ′ − ( − ) − + = − − ( + 1) V. Applications of DE Logistic DE = 1+ − where = −( )
( ) and M = Carrying Capacity. Example: A population grows logistically where the maximum sustainable population is 1500. Assuming
that the initial population is 350 and after 5 years the population has grown to 500, find an expression for
the population in t years.
1500 − 350 23
=
=
350
7 = 1500
1+ 23 − 7 5 = 500 = 1500
1+ 23
7 −5 , solve for k 1
23 = ln
5
14
1500 ∴ =
1+ 23
7 1 23 −5 ln 14 Chapter 10: Parametric Equations
I. Curves defined by Parametric Equations
Parametric equations = () = () Example: Sketch the curve identified by the parametric equations = 2 − 2 and = + 1.We take
several values for t and determine x and y values, then sketch these values on an x-y axis.
t
-1
0 x
3
0 y
0
1 1
2
3 -1
0
3 2
3
4 II. Calculus with Parametric Equations
When given a curve defined by x and y, you should be able to show tangents to the curve at a given point,
the point where the tangent is vertical or horizontal, and find arc length.
III. Polar Coordinates
Equations to know: = , = ∴ 2 + 2 = 2 , = (, )
Convert from Cartesian to polar.
2 − 5 3 = 1 + 2 − 5 3 = 1 + 2 − 5 3 cos 3 = 1 + 2 Convert from polar to Cartesian.
−8 = −8 =
2 + 2 −8
= 2 + 2
2 + 2
−8 = 2 + 2 2, 3 to Cartesian: = = 2 = 1
3 = = 2 = 3
3
∴ (1, 3)
(1,-1) to polar:
= 2 + 2 = 1 2 + −1 2 = 2 = 1
= −1
7
7
=
∴ ( 2, )
4
4 Chapter 11: Sequences & Series
I. Sequences
A sequence has the form {a n } where n is a positive integer.
Example of a sequence: = 10+ You should be able to evaluate the limit of a sequence. lim → ∞ = . If L exists, {an} is convergent and if
L doesn’t exist, {an} is divergent.
Example: Find lim → ∞ +1 = lim →∞ 1
1+ 1 = 1The limit exists and the sequence is convergent.
II. Series
∞ =1 Divergence Test: If lim →∞ = ±∞ or if lim →∞ ≠ 0, then
0, it does not mean that the series is convergent! ∞
=1 is divergent. If lim →∞ = Example: Is the following series convergent or divergent?
∞ =1 2
3 + 1 2
→∞ 3 + 1
2
2
lim
1 = 3 ≠ 0 ∴ Divergent
→∞ 3 + lim Geometric Series: 1− =
,
1− if r < 1, = ∞ −1 =1 and the series is convergent; if r > 1, .
1− Example: Is the following series convergent or divergent? ∞
=1 5 2 −3 −1
2 In this series, = 5, = − 3 , < 1, so the series will be convergent.
= 5
=
= 3 ∴ convergent
1− 1+2
3
∞ =1 3 −1 We must first change this series into the form of a geometric series:
∞ −1
=
3 −1 =1 ∞ −1 −1 3 In this series, = , = 3 , < 1, so the series will be convergent.
= ∞ =1
∞ = 3
3− 3 1+2
3
( =1 =
1− 1− 1 2
+ )
3 3
1 ∞ =1 1
+
3 ∞ =1 2
3 ∞ =>
=1 1 1
3 3 2
∞ −1 +
=1 2 2
3 3 −1 1 1: = 1
= 3 1 = ∴ convergent
1− 1−
2
3
2 2: =
= 3 = 2 ∴ convergent
1− 1−2
3
1 + 2 =
III. Integral Test
∞ −
=0 2 1
5
+ 2 = ∴ the series is convergent.
2
2 = −
∞ 2 2 − evaluate using improper integrals.
0 2 − =
0 P-series: 1
∞
= 1
2 ∴ since f x converges, the series converges . If K>0, then the p-series converges for p>1 and diverges for p≤1. IV. Comparison Test
Is the following series convergent or divergent?
∞ =1 1
2 +1 1
1
< 2 +1 2
∞
1
can be evaluated using geometric series and is found to be convergent.
2
=1
∞ ∴
=1 1
is convergent.
2 +1 V. Alternating Series
∞ −1 −1 =1 Alternating Series Test: If +1 ≤ lim →∞ = 0, then the series converges.
∞ Example: Does the series −1 −1 2 converge or diverge? =1 2 +1 2 2 < and lim = 0 = 1
→∞ It does not satisfy the second condition, so the series diverges.
VI. Absolute Convergence & Ratio Test is absolutely convergent if
||is not convergent.
∞ is convergent but −1 −1 =1
∞ |
=1 ||is convergent. It is conditionally convergent if ∞ −1 −1 |=
=1 1 determine
2 using p−series convergent ∴ absolutely convergent. Ratio Test for Absolute Convergence = lim | →∞ +1
| Case L<1
L>1 or L=infinity
L=1 Series is absolutely convergent
Series is divergent
Inconclusive
∞ Example: Is the series =0 −1 onvergent or divergent?
5 + 1 −1 +1
5 +1 +1
−1 →∞
5 +1 = lim
= lim −1 →∞ = lim →∞ 5+
5+ 5 + 1
5 + 6 1 6 = 1 ∴ Inconclusive, we must use another method to find the answer (alternating series test. ) Root Test(applied when nth power occurs) = lim →∞ | | Analyze L based on previous table.
∞ Example: Is the series
=1 2 + 1
22 + 1 convergent or divergent? 1 2 + 1
= lim
→∞
22 + 1
2 + 1
= lim 2
→∞ 2 + 1
1
= ∴ converges, absolute convergence
2 VII. Strategy for Testing Series
When asked to evaluate the convergence or divergence of a series, look for…
1. ∞ =1 1/ = 2. ∞
=1 P-series −1 = Geometric Series 3. If a series has a form similar to p-series or geometric series, consider using the comparison test.
4. If you can see at a glance that lim →∞ ≠ 0, then Divergence Test should be used. 5. ∞
=1 −1 −1 ∞
=1 −1 = alternating series tests 6. ( ) = root test
7. If = , where ∞
1 is easily evaluated = integral test 8. Else = Absolute Convergence Test
VIII. Power Series
∞ =0 Radius of Convergence: There is a positive number R such that the series converges if |x-a|<R and
diverges if |x-a|>R.
∞ Example: For which values of x is
=1
+1 We use the ratio test: lim →∞ −3 −3
+1 −3 convergent and what is the radius of convergence? −3 →∞
+1 = lim
−3
→ ∞ + 1
= −3
By the ratio test, the series is convergent when |x-3|<1 and divergent when |x-3|>1.Radius of
convergence = 1. Now we must determine the intervals of x for which the series is convergent.
= lim . . − 3 < 1 −1 < − 3 < 1
2< <4
∴ The series converges when 2<x<4 and diverges when x<2 and x>4. Now we must determine what
happens when |x-3| = 1 (x=2, x=4.) When x=4, the series becomes
When x=2, the series becomes −1 1 , evaluate using p-series, divergent. , which converges by alternating series test. ∴ The series converges for 2 ≤ < 4.
IX. Representation of Functions as Power Series
∞ =
=0 Example: Express
= 1 <1
1− 1
as the sum of a power series and find the interval of convergence.
1 + 2 1
1 − − 2
∞ − 2 =
=0 evaluate using geometric series and we find the interval of convergence to be −1,1 . Example: Express = −1 2 as a power series.
2
tan−1 2 + = 1 + 4 2
∞
1
=2 = 2
−4 2 1 − −4 2
=0 ∞ =2 −4 2 ∞ 2 =0 =0 = 0, tan−1 2 0
tan 2 −1 2 −1 22+1 2 +1
2 + 1 =0 = 2 +1
+ 2 + 1 2 +1
2 + 1 2 = 2
∞ = 0, = 0 ∞
−1 −4 =0 X. Taylor & Maclaurin Series
∞ : = =0
∞ ! : = 0:
=0 − 0
! = + ′ − + ′′ 2! − 2 +⋯ Example: Find the Maclaurin series for = . = for any value of n so 0 = 1
∞ =
=0 1 =
! ∞ =0 ! Example: Find the Taylor Series for = = 2.
∞ =0 ∞ 2
! −2 = =0 2
−2
! Taylor’s Inequality (Remainder Theorem) ~ Used to estimate error
| | = Example: Approximate the function =
is this approximation when 7 ≤ ≤ 9. | − | +1
, ≥ +1 ()
+1 !
3 by a Taylor polynomial of degree 2 at a=8. How accurate ′ 2 = 8 + 8 ′′ 8
−8 +
2! −8 2 2 () = 2 + 1
1
−
−8
12 − 8
288 2 = −8
3! 2 3 ≥ 3 . Now we must find M using either 7 or 9. We want to use the x value that will maximize M.
For x=9, 3 9 = 10
7 1
8 < 0.0011 9 3 10 1
< 0.0021
7 78
3
0.0021
∴ = 0.0021 and 2 =
7−8
3!
For = 7, 3 7 = 3 = 0.00035 ...

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