Mean = 124.9
Median = 128
Standard deviation = 18.3
Problem 1
(a) Set u=4x and write 1/sin u as csc u. Then the answer is (-1/4)ln |csc 4x + cot 4x| + C.
(b) Multiply numerator and denominator by (1-cos x), so the denominator becomes sin^2 x. The integral splits into three pieces. The first is csc^2 x which integrates to -cot x. The second is -2cos x/sin^2 x. Let u = sin x, so du = cos x dx, giving 2/sin x. The last is cos^2 x/sin^2 x = csc^2 x - 1. This integrates to -cot x - x. The answer is then 2csc x - 2cot x - x + C = 2(1-cos x)/sin x - x + C.
Alternatively, 1-cos x = 2sin^2(x/2) and 1+cos x = 2cos^2(x/2). So the quotient is tan^2(x/2)=sec^2(x/2)-1, which integrates to 2tan(x/2)-x+C (which is the same as the previous answer through the tan of half-angle trig identity).
Many people missed the second part of this. Some attempted the tan x/2 substitution (which will work eventually).
Problem 2
Using the disk method, the answer is the integral of pi(ln x)^2 between 1 and 2. By parts, let u=(ln x)^2 and dv = pi dx. Then du = (2ln x)/x dx and v = pi x. Then we need to integrate 2pi ln x = 2pi x ln x - 2pi x. Putting it together, we get pi x (ln x)^2 - 2pi x ln x + 2pi x, evaluated between 1 and 2, to get 2pi(1-ln 2)^2, or about 0.59.
Several tried the Theorem of Pappus, which will work. Some tried the shell method, but set it up incorrectly (15 point deduction). Some missed the subtraction of 1 in the integral evaluation (-4 points).
Problem 3
Let (X,Y) be this centroid. The parabola has x-intercepts at x=0 and x=2 and is symmetric about the line x=1. Its maximum is achieved at x=1, and is 1. Symmetry allows us to say that X=1. The area is the integral of 2x-x^2 between 0 and 2, which is 4/3. The easiest way to compute the y-coordinate of the centroid is to use the Theorem of Pappus. 2pi Y is the volume of rotation divided by the area. The volume is computed by integrating pi y^2 dx from 0 to 2. This volume is 16pi/15, so when divided by 2pi (4/3) gives 2/5. So the centroid is (1, 2/5).
Alternatively, one could compute the moment about the x-axis. We must solve the parabolic equation for x: x = 1+sqrt(1-y) and 1-sqrt(1-y). The region lies between these two curves, so the moment is calculated by integrating the difference times y: 2y sqrt(1-y), between 0 and 1.
I awarded 10 points for the area and 10 points for each coordinate. There were many calculation mistakes, including missing factors of (1/2) or 2, which usually resulted in a 2-6 point deduction.
Problem 4
(a) Using L'Hospital's rule once gives ln 2/ln 5.
(b) This has the form (infinity)^0. Taking logs gives ln(1+x)/e^x, which is (infinity)/(infinity). Now apply L'Hospital: 1/((1+x)e^x), which clearly goes to 0 as x goes to infinity. Exponentiating gives e^0=1. I deducted 4 points for not exponentiating the 0 at the end.
(c) Although L'Hospital can be applied, this is more easily done by writing it as 1/ln x + 1/x. Both limits are 0, so the sum is 0.
If one applies L'Hospital, the resulting form is (1+1/x)/(1+ln x). This clearly tends to 0, but some fell into the L'Hospital trap and differentiated again. That was a 4 point deduction. Others, instead of simply pointing out that this tends to 0, simplified to (x+1)/(x+xln x), which is another indeterminate form, and requires another application of L'Hospital (which does solve the problem).
(d) L'Hospital does not apply since the denominator is not 0 at x=3. The limit is 0.
Problem 5
(a) Write 2^x 3^(2x) as 18^x=e^(x ln 18). Now integrate and evaluate to get (18/ln 18) (18^2-1) = 2011.5
(b) Let u = ln ln x, so du = 1/(x ln x) dx. Then this integrates to u^2/2 or (ln ln x)^2/2 + C.
(c) x^(-4/3) integrates to -3/x^(1/3). Thus the definite integral is 3-3/t^(1/3), which tends to 3 as t goes to infinity.
Only the first part caused difficulties. A surprising fraction of the class still has difficulty manipulating exponential forms. Such incorrect manipulations resulted in significant point loss. A few integrated by parts - a heavy hammer indeed. That approach will eventually work, but is gritty
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