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Math 113 Calculus II Final Exam Form A KEY
Introduction
This document contains the answer key and detailed solutions for the Math 113 Calculus II Final Exam, Form A. The exam covers fundamental concepts of integral calculus, techniques of integration, sequences and series, and applications of integration. Each problem includes step-by-step solutions to help understand the correct approach and methodology.
Problem 1: Definite Integrals
Problem: Evaluate the definite integral (3x + 2x - 1) dx.
Solution:
Using the power rule for integration:
(3x + 2x - 1) dx = x + x - x + C
Evaluating from 0 to 2:
[x + x - x] = (8 + 4 - 2) - (0 + 0 - 0) = 10
Therefore, the value of the integral is 10.
Using the power rule for integration:
(3x + 2x - 1) dx = x + x - x + C
Evaluating from 0 to 2:
[x + x - x] = (8 + 4 - 2) - (0 + 0 - 0) = 10
Therefore, the value of the integral is 10.
Problem 2: Integration by Parts
Problem: Evaluate xln(x) dx.
Solution:
Using integration by parts with u = ln(x) and dv = x dx:
du = (1/x) dx and v = x/2
xln(x) dx = (x/2)ln(x) - (x/2)(1/x) dx
= (x/2)ln(x) - (x/2) dx
= (x/2)ln(x) - x/4 + C
= (x/4)[2ln(x) - 1] + C
Using integration by parts with u = ln(x) and dv = x dx:
du = (1/x) dx and v = x/2
xln(x) dx = (x/2)ln(x) - (x/2)(1/x) dx
= (x/2)ln(x) - (x/2) dx
= (x/2)ln(x) - x/4 + C
= (x/4)[2ln(x) - 1] + C
Integration by Parts Formula: u dv = uv - v du
Problem 3: Trigonometric Integrals
Problem: Evaluate sin(x) dx.
Solution:
Using the identity sin(x) = 1 - cos(x):
sin(x) dx = sin(x)sin(x) dx
= sin(x)[1 - cos(x)] dx
Let u = cos(x), then du = -sin(x) dx:
= -(1 - u) du
= - (1 - u) du
= -[u - u/3] + C
= -cos(x) + cos(x)/3 + C
= cos(x)/3 - cos(x) + C
Using the identity sin(x) = 1 - cos(x):
sin(x) dx = sin(x)sin(x) dx
= sin(x)[1 - cos(x)] dx
Let u = cos(x), then du = -sin(x) dx:
= -(1 - u) du
= - (1 - u) du
= -[u - u/3] + C
= -cos(x) + cos(x)/3 + C
= cos(x)/3 - cos(x) + C
Problem 4: Partial Fractions
Problem: Evaluate (5x - 3)/(x - 2x - 3) dx.
Solution:
First factor the denominator: x - 2x - 3 = (x - 3)(x + 1)
Using partial fractions: (5x - 3)/[(x - 3)(x + 1)] = A/(x-3) + B/(x+1)
5x - 3 = A(x + 1) + B(x - 3)
Setting x = 1: 2 = 4A, so A = 1/2
Setting x = -1: -8 = -4B, so B = 2
[(1/2)/(x - 3) + 2/(x + 1)] dx
= (1/2)ln|x - 3| + 2ln|x + 1| + C
= (1/2)[ln|x - 3| + 4ln|x + 1|] + C
= (1/2)ln|(x - 3)(x + 1)| + C
First factor the denominator: x - 2x - 3 = (x - 3)(x + 1)
Using partial fractions: (5x - 3)/[(x - 3)(x + 1)] = A/(x-3) + B/(x+1)
5x - 3 = A(x + 1) + B(x - 3)
Setting x = 1: 2 = 4A, so A = 1/2
Setting x = -1: -8 = -4B, so B = 2
[(1/2)/(x - 3) + 2/(x + 1)] dx
= (1/2)ln|x - 3| + 2ln|x + 1| + C
= (1/2)[ln|x - 3| + 4ln|x + 1|] + C
= (1/2)ln|(x - 3)(x + 1)| + C
Problem 5: Improper Integrals
Problem: Determine if the improper integral ^ (1/x) dx converges, and if so, evaluate it.
Solution:
^ (1/x) dx = lim(b) ^b (1/x) dx
= lim(b) [-1/x]^b
= lim(b) [-(1/b) + 1]
= 1
Since the limit exists and is finite, the improper integral converges to 1.
^ (1/x) dx = lim(b) ^b (1/x) dx
= lim(b) [-1/x]^b
= lim(b) [-(1/b) + 1]
= 1
Since the limit exists and is finite, the improper integral converges to 1.
Problem 6: Sequences
Problem: Determine if the sequence a = (2n + 3)/(5n - 1) converges, and if so, find its limit.
Solution:
To find the limit as n approaches infinity:
lim(n) (2n + 3)/(5n - 1)
Dividing numerator and denominator by n:
= lim(n) (2 + 3/n)/(5 - 1/n)
= (2 + 0)/(5 - 0)
= 2/5
Therefore, the sequence converges to 2/5.
To find the limit as n approaches infinity:
lim(n) (2n + 3)/(5n - 1)
Dividing numerator and denominator by n:
= lim(n) (2 + 3/n)/(5 - 1/n)
= (2 + 0)/(5 - 0)
= 2/5
Therefore, the sequence converges to 2/5.
Problem 7: Series Tests
Problem: Determine if the series (n=1 to ) 1/n converges or diverges.
Solution:
Using the p-series test, with p = 2 > 1:
The series (1/n) converges because p > 1.
This is a specific case of the Basel problem, and it converges to /6.
Using the p-series test, with p = 2 > 1:
The series (1/n) converges because p > 1.
This is a specific case of the Basel problem, and it converges to /6.
P-Series Test: (1/n) converges if p > 1 and diverges if p 1
Problem 8: Ratio Test for Series
Problem: Use the ratio test to determine if the series (n=1 to ) (2)/(n!) converges or diverges.
Solution:
Using the ratio test:
lim(n) |a/a| = lim(n) |(2/((n+1)!))/(2/n!)|
= lim(n) |(2/n!)/(2/(n+1)!)|
= lim(n) |(2(n+1)!)/(2n!)|
= lim(n) |(2(n+1))/(1)|
= lim(n) (2/(n+1))
= 0
Since 0 < 1, the series converges absolutely by the ratio test.
Using the ratio test:
lim(n) |a/a| = lim(n) |(2/((n+1)!))/(2/n!)|
= lim(n) |(2/n!)/(2/(n+1)!)|
= lim(n) |(2(n+1)!)/(2n!)|
= lim(n) |(2(n+1))/(1)|
= lim(n) (2/(n+1))
= 0
Since 0 < 1, the series converges absolutely by the ratio test.
Ratio Test: If lim(n) |a/a| = L, then: - If L < 1, the series converges absolutely - If L > 1, the series diverges - If L = 1, the test is inconclusive
Problem 9: Power Series
Problem: Find the radius and interval of convergence for the power series (n=0 to ) ((x-3))/(n+1).
Solution:
Using the ratio test:
lim(n) |((x-3)/(n+2))/((x-3)/(n+1))|
= lim(n) |(x-3)(n+1)/(n+2)|
= |x-3|
For convergence, we need |x-3| < 1, i.e., -1 < x-3 < 1, or 2 < x < 4.
So the radius of convergence is 1.
Checking endpoints:
For x = 2: ((-1))/(n+1) converges by the alternating series test
For x = 4: (1)/(n+1) diverges by the harmonic series
Therefore, the interval of convergence is [2, 4).
Using the ratio test:
lim(n) |((x-3)/(n+2))/((x-3)/(n+1))|
= lim(n) |(x-3)(n+1)/(n+2)|
= |x-3|
For convergence, we need |x-3| < 1, i.e., -1 < x-3 < 1, or 2 < x < 4.
So the radius of convergence is 1.
Checking endpoints:
For x = 2: ((-1))/(n+1) converges by the alternating series test
For x = 4: (1)/(n+1) diverges by the harmonic series
Therefore, the interval of convergence is [2, 4).
Problem 10: Taylor Series
Problem: Find the Taylor series for f(x) = e^x about x = 0 (the Maclaurin series) and determine its interval of convergence.
Solution:
The derivatives of f(x) = e^x are f(x) = e^x for all n.
f(0) = e^0 = 1 for all n.
Therefore, the Maclaurin series is:
f(x) = (n=0 to ) ((f(0))/(n!))x
e^x = (n=0 to ) (1/n!)x
= 1 + x + x/2! + x/3! + x/4! + ...
To find the interval of convergence:
Using the ratio test:
lim(n) |(x/(n+1)!)/(x/n!)|
= lim(n) |x/(n+1)|
= 0 for all x
Since 0 < 1 for all x, the series converges for all real numbers, so the interval of convergence is (-, ).
The derivatives of f(x) = e^x are f(x) = e^x for all n.
f(0) = e^0 = 1 for all n.
Therefore, the Maclaurin series is:
f(x) = (n=0 to ) ((f(0))/(n!))x
e^x = (n=0 to ) (1/n!)x
= 1 + x + x/2! + x/3! + x/4! + ...
To find the interval of convergence:
Using the ratio test:
lim(n) |(x/(n+1)!)/(x/n!)|
= lim(n) |x/(n+1)|
= 0 for all x
Since 0 < 1 for all x, the series converges for all real numbers, so the interval of convergence is (-, ).
Taylor Series Formula: f(x) = (n=0 to ) ((f(a))/(n!))(x-a)
Problem 11: Applications of Integration - Area
Problem: Find the area of the region bounded by y = x and y = x.
Solution:
First, find the intersection points:
x = x
x = x
x - x = 0
x(x - 1) = 0
x = 0 or x = 1
On the interval [0, 1], x x.
Area = (x - x) dx
= (x/ - x) dx
= [(x/)/(3/2) - x/3]
= [2/3x/ - x/3]
= [2/31 - 1/3] - [0 - 0]
= 2/3 - 1/3
= 1/3 square units
First, find the intersection points:
x = x
x = x
x - x = 0
x(x - 1) = 0
x = 0 or x = 1
On the interval [0, 1], x x.
Area = (x - x) dx
= (x/ - x) dx
= [(x/)/(3/2) - x/3]
= [2/3x/ - x/3]
= [2/31 - 1/3] - [0 - 0]
= 2/3 - 1/3
= 1/3 square units
Problem 12: Volume of Solids of Revolution
Problem: Find the volume of the solid generated by revolving the region bounded by y = x, the x-axis, and x = 2 about the y-axis.
Solution:
Using the shell method:
Volume = 2 (radius)(height) dx
= 2 x(x) dx
= 2 x dx
= 2[x/4]
= 2[16/4 - 0]
= 24
= 8 cubic units
Using the shell method:
Volume = 2 (radius)(height) dx
= 2 x(x) dx
= 2 x dx
= 2[x/4]
= 2[16/4 - 0]
= 24
= 8 cubic units
Shell Method Formula: Volume = 2[a,b] (radius)(height) dx
Problem 13: Arc Length
Problem: Find the length of the curve y = ln(cos(x)) from x = 0 to x = /4.
Solution:
The formula for arc length is:
L = [a,b] (1 + (dy/dx)) dx
First, find dy/dx:
dy/dx = -(sin(x))/(cos(x)) = -tan(x)
(dy/dx) = tan(x)
L = ^(/4) (1 + tan(x)) dx
= ^(/4) (sec(x)) dx
= ^(/4) |sec(x)| dx
= ^(/4) sec(x) dx (since sec(x) is positive on [0, /4])
= [ln|sec(x) + tan(x)|]^(/4)
= ln|sec(/4) + tan(/4)| - ln|sec(0) + tan(0)|
= ln|2 + 1| - ln|1 + 0|
= ln(1 + 2)
Therefore, the length of the curve is ln(1 + 2) units.
The formula for arc length is:
L = [a,b] (1 + (dy/dx)) dx
First, find dy/dx:
dy/dx = -(sin(x))/(cos(x)) = -tan(x)
(dy/dx) = tan(x)
L = ^(/4) (1 + tan(x)) dx
= ^(/4) (sec(x)) dx
= ^(/4) |sec(x)| dx
= ^(/4) sec(x) dx (since sec(x) is positive on [0, /4])
= [ln|sec(x) + tan(x)|]^(/4)
= ln|sec(/4) + tan(/4)| - ln|sec(0) + tan(0)|
= ln|2 + 1| - ln|1 + 0|
= ln(1 + 2)
Therefore, the length of the curve is ln(1 + 2) units.
Arc Length Formula: L = [a,b] (1 + (dy/dx)) dx
Conclusion
This answer key provides detailed solutions to the problems on the Math 113 Calculus II Final Exam, Form A. Students are encouraged to compare their solutions with those provided here to identify areas for further study. Remember that the key to success in calculus lies not only in memorizing formulas but also in understanding the underlying concepts and developing problem-solving strategies that can be applied to a variety of situations.
For additional practice, consider exploring similar problems from your textbook or past exams. Working through a diverse range of problems will strengthen your understanding of these fundamental calculus concepts and prepare you for future mathematics courses.
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