Math 1b Calculus II Final Exam Solutions
Calculus II is a challenging course that builds upon the fundamental concepts introduced in Calculus I. This detailed guide provides solutions to common types of problems that you might encounter on your Math 1b Calculus II final exam. Mastering these techniques will help you approach the final with confidence.
Common Topics Covered
Before diving into specific solutions, it's important to understand the key topics typically covered on a Calculus II final exam:
- Integration techniques (substitution, integration by parts, partial fractions)
- Applications of integration (areas, volumes, arc length, surface area)
- Sequences and series (convergence tests, power series, Taylor series)
- Differential equations (separable, linear, homogeneous)
- Parametric equations and polar coordinates
Integration Techniques
Example 1: Integration by Parts
Find the integral xsin(x)dx
Solution:
Using integration by parts formula: udv = uv - vdu
Let u = x, which means du = dx
Let dv = sin(x)dx, which means v = -cos(x)
xsin(x)dx = x(-cos(x)) - (-cos(x))dx
= -xcos(x) + cos(x)dx
= -xcos(x) + sin(x) + C
Example 2: Trigonometric Substitution
Find the integral (x/((9-x)))dx
Solution:
Let x = 3sin(), where -/2 /2
Then dx = 3cos()d
(9-x) = (9-9sin()) = (9cos()) = 3|cos()| = 3cos() (since cos() 0 in this range)
Substituting:
(9sin()/(3cos()))3cos()d = 9sin()d
Using the identity sin() = (1-cos(2))/2:
= 9(1-cos(2))/2 d = (9/2)(1-cos(2))d
= (9/2)( - (1/2)sin(2)) + C
= (9/2)( - sin()cos()) + C
Since x = 3sin(), we have sin() = x/3 and cos() = (9-x)/3
Also, = arcsin(x/3)
= (9/2)(arcsin(x/3) - (x/3)((9-x)/3)) + C
= (9/2)arcsin(x/3) - (x(9-x))/2 + C
Example 3: Partial Fractions
Find the integral (x/(x-5x+6))dx
Solution:
First, factor the denominator: x-5x+6 = (x-2)(x-3)
Express the integrand using partial fractions:
x/[(x-2)(x-3)] = A/(x-2) + B/(x-3)
x = A(x-3) + B(x-2) = (A+B)x - 3A - 2B
Comparing coefficients: A+B = 1 and -3A-2B = 0
Solving this system: A = -2, B = 3
Therefore, (x/(x-5x+6))dx = (-2/(x-2) + 3/(x-3))dx
= -2ln|x-2| + 3ln|x-3| + C
Applications of Integration
Example 4: Finding the Area Between Curves
Find the area bounded by y = x and y = x
Solution:
First, find the points of intersection by setting the functions equal:
x = x x-x = 0 x(x-1) = 0 x = 0 or x = 1
For 0 x 1, we have x x.
The area = (x - x) dx
= [x/2 - x/3]
= (1/2 - 1/3) - (0 - 0)
= 1/6 square units
Example 5: Volume by Disk Method
Find the volume of the solid obtained by rotating the region bounded by y = x, x = 0, and x = 4 about the x-axis
Solution:
Using the disk method, the volume V = [f(x)] dx
In this case, f(x) = x, a = 0, and b = 4
V = (x) dx = x dx
= [x/2]
= (16/2 - 0)
= 8 cubic units
Sequences and Series
Example 6: Determining Series Convergence
Determine whether the series (n=1 to ) (n/2) converges or diverges
Solution:
We'll use the Ratio Test:
lim(n) |a/a| = lim(n) |((n+1)/2^(n+1))/(n/2)|
= lim(n) |((n+1)/2^(n+1))(2/n)|
= lim(n) |((n+1)/(2n))|
= lim(n) |(n(1+1/n))/(2n)|
= lim(n) |(1+1/n)/2| = 1/2
Since 1/2 < 1, the Ratio Test tells us that the series converges absolutely.
Example 7: Finding Power Series Radius of Convergence
Find the radius of convergence for the power series (n=0 to ) n(x-2)
Solution:
Using the Ratio Test:
lim(n) |a(x-2)^(n+1)/a(x-2)| = lim(n) |(n+1)(x-2)^(n+1)/(n(x-2))|
= lim(n) |(n+1)(x-2)/n|
= lim(n) |(1+1/n)(x-2)| = |x-2|
For convergence, we require |x-2| < 1
Therefore, the radius of convergence is 1, centered at x = 2.
Example 8: Taylor Series Expansion
Find the Taylor series expansion of f(x) = ln(x) centered at x = 1
Solution:
First, find the derivatives of f(x) at x = 1:
f(x) = ln(x), f(1) = 0
f'(x) = 1/x = x^(-1), f'(1) = 1
f''(x) = -1/x = -x^(-2), f''(1) = -1
f'''(x) = 2/x = 2x^(-3), f'''(1) = 2
f(x) = -6/x = -6x^(-4), f(1) = -6
General pattern: f(1) = (-1)^(n-1)(n-1)! for n 1
The Taylor series is:
ln(x) = (n=0 to ) [f(1)/n!](x-1)
= 0 + (n=1 to ) [(-1)^(n-1)(n-1)!/n!](x-1)
= (n=1 to ) [(-1)^(n-1)/n](x-1)
= (x-1) - (x-1)/2 + (x-1)/3 - (x-1)/4 + ...
This series converges for 0 < x < 2.
Differential Equations
Example 9: Separable Differential Equation
Solve the differential equation dy/dx = (x+1)/(xy) with y(1) = 2
Solution:
First, separate the variables:
dy/dx = (x+1)/(xy)
y dy = (x+1)/x dx
y dy = (x + 1/x) dx
Integrate both sides:
y dy = (x + 1/x) dx
y/2 = x/2 + ln|x| + C
Using the initial condition y(1) = 2:
2/2 = 1/2 + ln|1| + C
2 = 1/2 + 0 + C, so C = 3/2
Therefore, y/2 = x/2 + ln|x| + 3/2
y = x + 2ln|x| + 3
y = (x + 2ln|x| + 3)
Since y(1) = 2 > 0, we take the positive branch:
y = (x + 2ln|x| + 3)
Example 10: Linear Differential Equation
Solve the differential equation dy/dx + 2xy = x
Solution:
This is a linear first-order differential equation of the form dy/dx + P(x)y = Q(x)
Here, P(x) = 2x and Q(x) = x
The integrating factor is:
(x) = e^(P(x)dx) = e^(2x dx) = e^(x)
Multiply both sides by (x):
e^(x)dy/dx + 2xe^(x)y = xe^(x)
The left side is the derivative of ye^(x):
d/dx[ye^(x)] = xe^(x)
Integrate both sides:
ye^(x) = xe^(x) dx
Let u = x, so du = 2x dx, and x dx = du/2:
= (1/2)e^u du = (1/2)e^u + C = (1/2)e^(x) + C
ye^(x) = (1/2)e^(x) + C
y = (1/2) + Ce^(-x)
Tips for the Calculus II Final Exam
- Practice makes perfect: Work through as many practice problems as possible before the exam.
- Master the basics: Make sure you have a solid understanding of fundamental integration techniques before tackling complex problems.
- Know the convergence tests: For series questions, be familiar with all convergence tests and when to use each one.
- Check your work: Always verify your answers by differentiating or integrating backwards when possible.
- Manage your time: Don't spend too much time on a single problem. If you're stuck, move on and come back later.
- Watch for integration constants: Don't forget to add "+ C" to your indefinite integrals.
Conclusion
Success on your Math 1b Calculus II final exam requires a solid understanding of the concepts covered throughout the course. By reviewing the examples provided above and practicing similar problems, you'll build the confidence and skills needed to excel on the final. Remember to focus on understanding the underlying principles rather than simply memorizing formulas. Good luck with your Calculus II final exam!
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