MATH 166: HONORS CALCULUS II EXAM I SOLUTIONS
Introduction
Welcome to the solutions for MATH 166: Honors Calculus II Exam I. This document provides detailed solutions to problems that appeared on the exam, along with explanations of key concepts. The exam covered integration by parts, trigonometric integrals, substitution, partial fractions, and applications of definite integrals. Each solution shows the step-by-step approach to solving the problem.
Problem 1: Integration by Parts
Evaluate: xe dx
Solution: Use integration by parts: udv = uv - vdu
Let u = x (so du = dx) and dv = e dx (so v = (1/3)e)
xe dx = x(1/3)e - (1/3)e dx
= (1/3)xe - (1/9)e + C
= (1/3)e(x - 1/3) + C
Problem 2: Trigonometric Integrals
Evaluate: sin(x)cos(x) dx
Solution: Separate one factor of sin(x):
sin(x)cos(x) = sin(x)cos(x)sin(x)
Using sin(x) = 1 - cos(x):
= (1 - cos(x))cos(x)sin(x)
= cos(x)sin(x) - cos(x)sin(x)
Let u = cos(x), du = -sin(x) dx:
= - u du + u du
= -(1/3)u + (1/5)u + C
= -(1/3)cos(x) + (1/5)cos(x) + C
Problem 3: Trigonometric Substitution
Evaluate: dx/(x(x + 9))
Solution: Use substitution: x = 3tan(), dx = 3sec() d
This gives (x + 9) = 3sec()
dx/(x(x + 9)) = (3sec() d)/(9tan()3sec())
= (1/9) cos()/sin() d
Let u = sin(), du = cos() d:
= (1/9) (1/u) du
= -(1/9)(1/u) + C
= -(1/9)(1/sin()) + C
From our substitution, sin() = x/(x + 9):
= -(1/9)((x + 9)/x) + C
Problem 4: Partial Fractions
Evaluate: (5x + 3)/(x + x) dx
Solution: Factor the denominator: x + x = x(x + 1)
Express as sum of simpler fractions:
(5x + 3)/(x(x + 1)) = A/x + B/(x + 1)
5x + 3 = A(x + 1) + Bx = (A + B)x + A
Comparing coefficients: A + B = 5, A = 3
Therefore, B = 2
(5x + 3)/(x + x) dx = 3/x dx + 2/(x + 1) dx
= 3 ln|x| + 2 ln|x + 1| + C
= ln|x(x + 1)| + C
Problem 5: Area Between Curves
Find the area bounded by y = x and y = x + 2.
Solution: Find intersection points where x = x + 2:
x - x - 2 = 0
(x - 2)(x + 1) = 0
So the curves intersect at x = -1 and x = 2.
For -1 x 2, the line y = x + 2 lies above the parabola y = x.
A = from -1 to 2 (x + 2 - x) dx
= [x/2 + 2x - x/3] from -1 to 2
= (2 + 4 - 8/3) - (1/2 - 2 + 1/3)
= (10/3) - (-7/6)
= 27/6
= 9/2
The area is 9/2 square units.
Problem 6: Volume of Revolution
Find the volume of the solid generated by revolving y = (x), the x-axis, and x = 4 about the x-axis.
Solution: Use the disk method for volume of revolution:
V = from a to b [f(x)] dx
In this case, f(x) = (x), a = 0, b = 4:
V = from 0 to 4 [(x)] dx
= from 0 to 4 x dx
= [x/2] from 0 to 4
= (16/2)
= 8
The volume is 8 cubic units.
Problem 7: Improper Integral
Determine whether from 1 to (1/x) dx converges or diverges.
Solution: An improper integral with an infinite upper limit is evaluated using a limit:
from 1 to (1/x) dx = lim as t from 1 to t (1/x) dx
= lim as t [-1/x] from 1 to t
= lim as t [(-1/t) - (-1/1)]
= lim as t [(-1/t) + 1]
= 0 + 1
= 1
Since the limit exists and is finite, the improper integral converges to 1.
Problem 8: Work Done Against Gravity
A 50-foot chain weighing 5 pounds per foot hangs from a winch. Find the work required to wind the entire chain onto the winch.
Solution: Consider a small segment of the chain of length y located at a distance y from the top of the chain.
This segment weighs 5y pounds and must be lifted approximately y feet.
The work done on this segment is approximately 5yy foot-pounds.
Total work = from 0 to 50 5y dy
= 5 [y/2] from 0 to 50
= 5 (2500/2)
= 6250
The work required is 6250 foot-pounds.
Problem 9: Arc Length
Find the length of the curve y = (x/6) + (1/(2x)) from x = 1 to x = 3.
Solution: Use the arc length formula: L = from a to b (1 + [f'(x)]) dx
First, find the derivative: f'(x) = (x/2) - (1/(2x))
Compute (1 + [f'(x)]):
1 + [f'(x)] = 1 + [(x/2) - (1/(2x))]
= 1 + (x/4) - (1/2) + (1/(4x))
= [(x/2) + (1/(2x))]
So (1 + [f'(x)]) = (x/2) + (1/(2x))
Now compute the arc length:
L = from 1 to 3 [(x/2) + (1/(2x))] dx
= (1/2)[(x/3) - (1/x)] from 1 to 3
= (1/2)[(27/3 - 1/3) - (1/3 - 1)]
= (1/2)[(26/3) - (-2/3)]
= (1/2)(28/3)
= 14/3
The curve length is 14/3 units.
Additional Tips for Calculus II
Mastering Honors Calculus II requires understanding techniques and their applications:
1. Practice identifying which integration technique is most appropriate for a given problem. Some problems may require a combination of techniques.
2. Pay careful attention to algebraic manipulation and simplification. A complex integral often becomes manageable after appropriate manipulation.
3. Draw diagrams to help visualize regions when working with applications of integrals.
4. Review trigonometric identities thoroughly, as they are essential for solving many integration problems.
5. Be careful with evaluation at bounds and signs when computing definite integrals.
6. Understand the connections between algebraic representations and geometric meanings of calculus concepts.
7. Practice problems regularly to build fluency with techniques and to recognize patterns in problem types.
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