Advanced Placement Calculus AB Practice

Section II Part A Solutions

A graphing calculator is required for some problems or parts of problems. You can verify your calculations online at SolveMyMath.com

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1. Let f be the functions given by:


Find the area of the region enclosed by the graphs of f(x) and -f(x).

Solution:
16 - (x - 1)² ≥ 0 and the domain of the function is [-3,5].
The answer can be found numerically by integrating f(x) - (-f(x)) between x = -3 and x = 5. However, an easier way to find the answer is to realize that f(x) is a semicircle of the circle (x - 1)² + y² = 16, y > 0, and -f(x) is the other semicircle, y < 0.
The radius of the circle is 4. The answer is ¶4² = 16¶.

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2. Let f and g be the functions given by:




Let R be the region in the first and second quadrants enclosed by the graphs of f and g. Find the area of R.

Solution: The answer can be found numerically by integrating f(x) - g(x) between their points of intersection. First we need to find these points:

4 - x² = .25x² + x + 1
1.25² + x - 3 = 0 with the solutions x = -2 and x = 1.2

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3. Let f and g be the functions given by



Let R be the region in the first and second quadrants enclosed by the graphs of f and g. Find the volume of the solid generated when R is revolved about the x axis.

Solution:

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4. Let f be a function derivabile on the (0, .6) interval. If f '(x) = 3 + ln(x + 1) and f(.1) = 2, what is f(.6)?.

Solution:

f(.6) = 3.647

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5. Let R be the region in the first quadrant bounded by the y axis and the graphs of f(x) = 3 + ln(x + 1) and g(x) = 2^3x. Find the area of R.

Solution:
3 + ln(x + 1) = 2^3x solved numerically gives x = .598.

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6. Let R be the region in the first quadrant bounded by the graphs of f(x) = 3 + ln(x + 1) and g(x) = 2^3x. Region R is the base of a solid. For this solid, each cross-section perpendicular to the x-axis is a square. Find the volume of this solid.

Solution:
3 + ln(x + 1) = 2^3x solved numerically gives x = .598.

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7. A checking account has $120,000 at time t=0. During the interval 0 ≤ t ≤ 18 months, money is deposited into the checking account at a rate d(t) = 2,000(t+1)sin(t/4) dollars/month. At the same time, money is withdrawn from the checking account at the rate w(t) = 10,000/(t+1) dollars/month. Is the dollar amount increasing at t=9 months? How many dollars are in the account at time t=18 months?

Solution:
d(9) - w(9) = 2000(9 + 1)sin(9/4) - 10000/(9 + 1)
d(9) - w(9) = 14,561 $/month is a positive value so money is deposited into the account at a higher rate than it is withdrawn at t = 9 months.
At t = 18 months, the sum available in the account is:

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