1. Let f(x) be the function on [0, Pi] whose graph is displayed below
The left, right, trapezoidal, and midpoint sums for L400, R400, T400 and M400 are, in some order, 6.095473, 6.107794, 6.120147, 6.107810. Identify which is which. Then determine if T400 underestimates or overestimates the underlying integral.
Solution: Since the curve is concave up the left sum will overestimate the integral, the right will underestimate it, the trapezoidal will overestimate it and, consequently the midpoint sum will underestimate the integral. The sums are R400 = 6.095473, M400 = 6.107794,
T400 = 6.107810, and L400 = 6.120147.
2. The functions f(x) = sin(x) + cos(x) and g(x) = cos(2x) on [0, Pi] are plotted below.
Using this evaluate
Solution: Since f(x) = sin(x) + cos(x) is an increasing function and g(x) = cos(2x) is a decreasing function for x close to 0, the curve for f(x) lies aboves the curve for g(x) at the start. They intersect at 3Pi/4. Consequently
3. Let y = f(x) be the function whose graph is
Let R be the region bounded by the curve y = f(x), the line x = 3, and the x-axis. Set up the integral for the volume generated by rotating R
a. around the line x = 4
b. around the line y = -2
c. around the y-axis
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a
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b
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c
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4. The conical tank sketched below, on the left, has a height of 2 ft and the circle at the top is of radius 2 ft. A vertical midsection is pictured on the right. The tank is filled with water, which weighs 62.5 lbs per cubic foot.
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How much work is done in pumping all of the water out of the cone to a point 3 foot above the top of the tank?
Solution: The volume of a disk lying y units above the x-axis is
Pi x2 dy. Its weight is 62.5 Pi x2 dy. Since it is to be lifted 3 foot above the top of the tank the work done in lifting it there is
(5-y) 62.5 Pi x2 dy. Thus the total work to be done is
Since y = x, (see the second graph) the total work is equal to
5. Graph the function 
Show and label all local minima and maxima. (You do not need to show the concavity).
on the6. The function f(x) is invertible domain [1,6], with inverse function g(y). Given the table
Solution: First, observe,
Next
so f'(x) = 0 for x = +1 and x = -1. By the limits above f(x) has local maximas at both x = -1 and x = +1.
Due to scaling problems, the main features of the graph has to be shown in two parts on the computer:
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6. Given the table
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x |
1 |
2 |
3 |
4 |
5 |
6 |
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f(x) |
-1 |
3 |
5 |
6 |
7 |
8 |
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f'(x) |
10 |
11 |
12 |
13 |
14 |
15 |
evaluate g'(5), where g is the inverse of f.
Soultion: Note first that f(3) = 5. Then g'(f(x)) = 1/f'(x) gives us g'(5) = 1/12.
7. Suppose that f(x) = x2 + x + 3a where a > 0 and the average value of f(x) on [0,a] is 13. What is the value of a?
Solution: We have
That is
or
