SPRING 2015 MAT 266 FINAL EXAM REVIEW
U-Substitution
U substitution comes in handy for breaking up integrals where one function is inside of another. For
example ∫(5 + 𝑥 2 )3 𝑑𝑥. In this example the inner function 5+x2 is inside the other function ( )3. To
solve:
1.
2.
3.
4.
let U be a variable to stand for the inner function ( in this case 5+x2)
Find the derivative of U with respect to x ( so dU = 2x dx in this case)
Substitute U for all x and dU for dx
Solve the integral, and if definite integral change all U back to its function of x and then
evaluate
Practice Problems Set 1
3𝑥 2
a) ∫ (8+𝑥3 )
b) ∫ cos(𝑥) ∗ (sin⁡(𝑥))3 𝑑𝑥
1
c) ∫ 𝑥(1+ln(𝑥))2 𝑑𝑥
Integration by Parts
Integration by parts is mainly used when we take the integral of the product of two functions. For
example ∫ 𝑥𝑒 𝑥 𝑑𝑥 cannot be integrated right away. But we can use
∫ 𝑢 ∗ 𝑣′ 𝑑𝑥 = 𝑢𝑣 − ∫ 𝑣 ∗ 𝑢′𝑑𝑥
1) determine⁡which⁡part⁡to⁡assign⁡to⁡u⁡and⁡which⁡to⁡assign⁡to⁡v’⁡in⁡the⁡above⁡property
a. Typically⁡you⁡will⁡want⁡to⁡pick⁡such⁡that⁡u’⁡vanishes⁡(in⁡the⁡example⁡we⁡pick⁡u=x⁡as⁡
u’=1)
2) Then find⁡v⁡and⁡u’⁡and⁡plug⁡them⁡into⁡the⁡format:⁡⁡𝑢𝑣 − ∫ 𝑣 ∗ 𝑢′𝑑𝑥
3) Solve the rest of the integral, possibly using another integration by parts for harder
problems
Practice Problems Set 2
a) ∫ 𝑥 𝑒 12𝑥 𝑑𝑥
b) ∫ 𝑥 2 𝑒 12𝑥 𝑑𝑥
c) ∫ 𝑥 cos⁡(𝑥)𝑑𝑥
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SPRING 2015 MAT 266 FINAL EXAM REVIEW
Trigonometric Substitutions
Remember these trig properties?
There are three cases where trig subs can help
If the integral contains √𝑎2 − 𝑥 2 then if we substitute in 𝑥 = a ∗ sin⁡(𝑢) then it simplifies to a*cos(u)
If the integral contains √𝑎2 + 𝑥 2 then if we substitute in 𝑥 = a ∗ tan(𝑢) then it simplifies to a*sec(u)
If the integral contains √𝑥 2 + 𝑎2 then if we substitute in 𝑥 = a ∗ sec(𝑢) then it simplifies to a*tan(u)
In these u is simply a variable like with U substitutions. You will still have to substitute in for x
everywhere
Practice Problems Set 3
Choose the best substitution for each of the following integrals
a) ∫
√4 2 −𝑥 2
𝑥
b) ∫
√5+𝑥 2
𝑥
c) ∫
√7+9𝑥 2
𝑥
Partial Fractions
𝑥+3
(𝑥+1)(𝑥−3)
Factor in Denominator
(𝑎𝑥 + 𝑏)
−0.5
6/4
+
𝑥+1
𝑥−3
Term in Partial Fraction Decomposition
𝐴
(𝑎𝑥+𝑏)
𝐴1
𝐴2
𝐴3
𝐴𝑘
+ (𝑎𝑥+𝑏)
2 + (𝑎𝑥+𝑏)3 + ⋯ + (𝑎𝑥+𝑏)𝑘
(𝑎𝑥+𝑏)
𝐴𝑥+𝐵
(𝑎𝑥 2 +𝑏𝑥+𝑐)
𝐴1 𝑥+𝐵1
𝐴 𝑥+𝐵2
𝐴𝑘 𝑥+𝐵𝑘
+ (𝑎𝑥 22+𝑏𝑥+𝑐)
2 + ⋯ + (𝑎𝑥 2 +𝑏𝑥+𝑐)𝑘
(𝑎𝑥 2 +𝑏𝑥+𝑐)
(𝑎𝑥 + 𝑏)𝑘
(𝑎𝑥 2 + 𝑏𝑥 + 𝑐)
(𝑎𝑥 2 + 𝑏𝑥 + 𝑐)𝑘
To find the coefficients of each of the factor’s term in the decomposition, set the original equal to the
sum of all the terms and then cross multiply to get an equation for the coefficients.
𝑥+3
𝐴
𝐵
∙ (𝑥 + 1)(𝑥 − 3) =
∙ (𝑥 + 1)(𝑥 − 3) +
∙ (𝑥 + 1)(𝑥 − 3)
(𝑥 + 1)(𝑥 − 3)
𝑥+1
𝑥−3
𝑥 + 3 = 𝐴(𝑥 − 3) + 𝐵(𝑥 + 1)
𝑥 = 𝐴𝑥 + 𝑏𝑥, 3 = −3𝐴 + 𝐵
You can also let x be a number that will simplify, for example let x= -1 or x=3 in turn and substitute in to
get: −1 + 3 = 𝐴(−1 − 3) + 𝐵(0), 3 + 3 = 𝐴(0) + 𝐵(3 + 1)
Practice Problems Set 4
Decompose the following partial fractions:
a)
20
(𝑥−4)(𝑥+1)
10𝑥 2 +14𝑥−36
b) ∫ (𝑥+3)(𝑥2 −2𝑥−3)
−4𝜋∙𝑥
c) ∫ (𝑥+1)2 (𝑥−1)
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SPRING 2015 MAT 266 FINAL EXAM REVIEW
Solutions to Sets 1-4
Set 1:
a) ln(8 + 𝑥 3 ) + 𝐶
1
b) 4 sin(𝑥)4 + 𝐶
1
c) − 2(1+ln(𝑥)) + 𝐶
Set 2:
1
12
1
(12 𝑥 2
1
) 𝑒 12𝑥 + 𝐶
144
1
1
− 144 𝑥 + 123 ) 𝑒 12𝑥
a) ( 𝑥 −
b)
+𝐶
c) 𝑥𝑠𝑖𝑛(𝑥) + cos(𝑥) + 𝐶
Set 3:
a) 4sin(t)
b) √5tan⁡(𝑡)
c)
√7
tan⁡(𝑡)
3
Set 4:
a)
−4
4
+
(𝑥+1)
(𝑥−4)
b) 5 ln(𝑥 + 1) + 4ln(𝑥 − 3) + ln(𝑥 + 3) + 𝐶
𝜋
c) 𝜋 ln(𝑥 + 1) + (𝑥+1) − 𝜋 ln(𝑥 − 1) + 𝐶
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SPRING 2015 MAT 266 FINAL EXAM REVIEW
Volumes
Volume by slicing
Disk Method
If S is a solid that lies between x=a and x=b and if its cross sectional area at a distance x can be written as
a function of x i.e. A(x), then the volume of the solid is given by
𝑏
V=∫𝑎 𝐴(𝑥)𝑑𝑥
Tip: Use this method when you can express a cross section of the volume as a function of a single
variable.
Washer method
When asked to find volume of a solid obtained by revolving the area enclosed between 2 curves,
Step 1) find point of intersection of the curves to get limits a and b.
Step 2) Express the area function of the cross section as
A=pi*((outer radius)2-(inner radius)2)
Step 3) Evaluate the integral as shown above
Volume by method of cylindrical shells
The Volume of a solid obtained by revolving the area under a curve y=f(x) between x=a and x=b
about the y axis can be obtained
𝑏
𝑉 = ∫ 2𝜋𝑥𝑓(𝑥)𝑑𝑥
𝑎
Use this method when you can’t easily express x in terms of y.
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SPRING 2015 MAT 266 FINAL EXAM REVIEW
Practice Problem Set 5
Applications to Physics
A) Work Done
If the force applied in doing a work for distance x can be expressed as f(x), then the work done in moving
from x=a to x=b is given by
𝑏
𝑊 = ∫ 𝑓(𝑥)𝑑𝑥
𝑎
Practice problem set 6
Sequences and Series
-A sequence is a list of numbers written in a definite order.
-A Series is a summation of terms of a sequence
-A sequence converges if lim 𝑎𝑛⁡ exists
𝑛→∞
-A sequence is called increasing if a1<a2<a3….. and decreasing if a1>a2>a3….
-A Series converges if lim 𝑆𝑛⁡ exists and is equal to some real value. Sn is the sum of first n terms of the
𝑛→∞
series
-A geometric series is convergent if |𝑟| < 1
-A series converges if lim 𝑎𝑛⁡ =0
𝑛→∞
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SPRING 2015 MAT 266 FINAL EXAM REVIEW
Ratio test
If lim |
𝑎𝑛+1
|=L<1
𝑎𝑛
then the series is absolutely convergent
If lim |
𝑎𝑛+1
|=L>1
𝑎𝑛
then the series is absolutely divergent
If lim |
𝑎𝑛+1
|=L=1
𝑎𝑛
then the test is inconclusive
𝑛→∞
⁡
𝑛→∞
⁡
𝑛→∞
⁡
Practice Problem set 7
Power series
-A series of the form Co +C1x+C2x2+C3x3……. Is called as a power series
-A series of the form Co +C1(x-a)+C2(x-a)2+C3(x-a)3……. Is called as a power series centered at a. This
series can converge in 3 possible waysa) if x=a or
b) for all x for certain functions
c) |𝑥 − 𝑎| < 𝑅 where R is the radius of convergence. The interval of convergence is all values of x for
which the series converges.
1
= 1 + 𝑥 + 𝑥2 + 𝑥3 … …
1−𝑥
Hence for any function that can be expressed as a modified version of the LHS above, the function can
be expressed as an infinite series similar to RHS.
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SPRING 2015 MAT 266 FINAL EXAM REVIEW
If a function can be expressed as an infinite series as shown above, differentiating or integrating the
function is equivalent to differentiating or integrating the infinite series.
Practice Problem set 8
Taylor series of f centered at a
If f(x) can be expressed in terms of a power series such as
∞
𝑓(𝑥) = ∑ 𝐶𝑛 (𝑥 − 𝑎)𝑛
𝑛=0
then 𝐶𝑛 =
𝑓𝑛 (𝑎)
𝑛!
When a=0, the series is called as a Maclaurin series.
Practice problem set 9
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SPRING 2015 MAT 266 FINAL EXAM REVIEW
Solutions to sets 5-9
Set 5
Set 6
2450 Joules
Set 7
Set 8
Set 9
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SPRING 2015 MAT 266 FINAL EXAM REVIEW
Parametric Curves
𝑥 = 𝑓(𝑡)⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡𝑦 = 𝑔(𝑡)
Each value of t determines a point (𝑥, 𝑦) that can be plotted in a coordinate plane. As t varies, the
point (x, y) = (f(t), g(t)) varies and traces out a curve C.
Tangent Lines to Parametric Curves
𝑑𝑦
𝑑𝑦 𝑑𝑡
𝑑𝑥
=
⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡𝑖𝑓⁡⁡ ⁡ ≠ 0
𝑑𝑥
𝑑𝑥
𝑑𝑡
𝑑𝑡
(𝑦 − 𝑦𝑖 ) = ⁡
𝑑𝑦
(𝑥 − 𝑥𝑖 )⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡𝑤ℎ𝑒𝑟𝑒⁡⁡⁡⁡𝑥𝑖 = 𝑓(𝑡𝑖 )⁡⁡⁡𝑦𝑖 = 𝑔(𝑡𝑖 )
𝑑𝑥
Practice Problems Set 10:
Areas of Regions Bounded by Parametric Curves
If the curve is given by the parametric equations 𝑥 = 𝑓(𝑡)⁡, 𝑦 = 𝑔(𝑡) and is traversed once as t
increases from 𝛼 to 𝛽
𝑏
𝛽
𝐴 = ⁡ ∫ 𝑦⁡𝑑𝑥 = ⁡ ∫ 𝑔(𝑡)𝑓 ′ (𝑡)𝑑𝑡
𝑎
𝛼
Practice Problem Set 11:
Lengths of Parametric Curves
If a curve C is described by parametric equations 𝑥 = 𝑓(𝑡)⁡, 𝑦 = 𝑔(𝑡) 𝛼 ≤ 𝑡 ≤ 𝛽 where⁡f’⁡and⁡g’⁡are⁡
continuous on the [𝛼, 𝛽] and C is traversed once as t goes from 𝛼 to 𝛽 then the length of C is:
𝛽
𝛽
𝑑𝑦/𝑑𝑡 2
𝑑𝑥 2
𝑑𝑌 2
𝐿 = ∫ √1 + (
) 𝑑𝑡 = ∫ √( ) + ( ) 𝑑𝑡
𝑑𝑥/𝑑𝑡
𝑑𝑡
𝑑𝑡
𝛼
𝛼
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SPRING 2015 MAT 266 FINAL EXAM REVIEW
Practice Problems Set 12:
Polar and Curves
𝑥 = 𝑟𝑐𝑜𝑠𝜃⁡⁡⁡⁡𝑦 = 𝑟𝑠𝑖𝑛𝜃⁡⁡⁡⁡⁡⁡𝑟 2 = 𝑥 2 + ⁡ 𝑦 2 ⁡⁡⁡⁡⁡⁡⁡⁡⁡𝜃 = 𝑎𝑟𝑐𝑡𝑎𝑛𝜃
𝑟 = 𝑓(𝜃)
𝑃𝑜𝑙𝑎𝑟⁡𝑐𝑢𝑟𝑣𝑒⁡𝑜𝑓⁡𝑟 = 𝑎⁡⁡𝑖𝑠⁡𝑐𝑖𝑟𝑐𝑙𝑒⁡𝑐𝑒𝑛𝑡𝑒𝑟𝑒𝑑⁡𝑎𝑡⁡𝑡ℎ𝑒⁡𝑜𝑟𝑖𝑔𝑖𝑛⁡𝑤𝑖𝑡ℎ⁡𝑟𝑎𝑑𝑖𝑢𝑠⁡𝑎⁡
𝑃𝑜𝑙𝑎𝑟⁡𝑐𝑢𝑟𝑣𝑒⁡𝑜𝑓⁡𝜃 = 𝐶⁡𝑖𝑠⁡𝑎⁡⁡𝑠𝑡𝑟𝑎𝑖𝑔ℎ𝑡⁡𝑙𝑖𝑛𝑒⁡𝑡ℎ𝑎𝑡⁡𝑝𝑎𝑠𝑠𝑒𝑠⁡𝑡ℎ𝑟𝑜𝑢𝑔ℎ⁡𝑂⁡𝑎𝑛𝑑⁡𝑚𝑎𝑘𝑒𝑠⁡𝑎𝑛⁡𝑎𝑛𝑔𝑙𝑒⁡𝐶⁡𝑤𝑖𝑡ℎ⁡𝑡ℎ𝑒⁡𝑝𝑜𝑙𝑎𝑟⁡𝑎𝑥𝑖𝑠⁡
Review of the Unit Circle:
Practice Problem Set 13 (Coordinate Conversions):
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SPRING 2015 MAT 266 FINAL EXAM REVIEW
Practice Problems Set 14 (Equation Conversions):
Tangents to Polar Curves
𝑑𝑦 𝑑𝑟
𝑑𝑦 𝑑𝜃 𝑑𝜃 𝑠𝑖𝑛𝜃 + 𝑟𝑐𝑜𝑠𝜃
=
=
𝑑𝑥 𝑑𝑥 𝑑𝑟 𝑐𝑜𝑠𝜃 − 𝑟𝑠𝑖𝑛𝜃
𝑑𝜃 𝑑𝜃
Horizontal Tangent:
𝑑𝑦
𝑑𝜃
=0
Vertical Tangent:
𝑑𝑥
𝑑𝜃
=0
Practice Problems Set 15:
Areas of Regions Bounded by Polar Curves
𝑏
𝐴 =⁡∫
𝑎
𝑏
1
1
[𝑓(𝜃)]2 𝑑𝜃⁡ = ⁡⁡∫ 𝑟 2 𝑑𝜃⁡
2
𝑎 2
Practice Problems Set 16:
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SPRING 2015 MAT 266 FINAL EXAM REVIEW
Solutions to Sets 10-16
Set 10:
Set 11:
Set 12:
Set 13:
Set 14:
Set 15:
Set 16:
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