Some Review Problems Exam 2

• Find y' at x = 1/2, for y = f(sin-1(x)), where f(x) is a function such that
f '(Pi/6) = 7.

By the chain rule,
y' = f '(sin -1(x))d(sin -1(x))/dx
= f '(sin -1(x))/(1 - x2)1/2. At x = 1/2 we get y' = f'(Pi/6)*2/31/2 = 7*2/31/2.

• Find the exact value of:

tan(sin-1(1/3)) = 1/81/2

• Simplify:

sin(sec-1(x)) = (x2 - 1)1/2/x

• Evaluate:
Use the substitutions u = sin(x), and cos2(x) = 1 - sin2(x) to get (sin3(x))/3 - (sin5(x))/5 + C.
• Evaluate using integration by parts:
Pick dv = e2xdx, u = x, to get (x/2)e2x - (1/4)e2x + C
• Evaluate using completing the square:
After completing the square, what we need to integrate is x/((x+1)2+1) dx. Now substitute u = x + 1 to get (u - 1)/(u2+1) du. This gives two integrals, u/(u2+1) du and -1/(u2+1) du. The first gives (1/2) ln |u2+1| + C, the second gives -arctan(u) + C. Thus the answer to the original problem is (1/2) ln |(x+1)2+1| - arctan(x+1) + C.
• Evaluate using partial fractions:
The answer is ln |x| + ln |x+2| + C.
• Determine whether the following improper integrals converge or diverge; justify your answers:

• This one converges: using the substitution u = x3 - 1, the integral is limt -> 1+ (2/3)(x3 - 1)1/2|t2 = (2/3)71/2 - 0.

This one converges, too: the integral is limR -> infinity arctan(x)|0R = Pi/2 - 0.
• Use a comparison to determine whether the following improper integral converges or diverges; completely justify your answer:
Since e-u is decreasing, and since x < x3/2 when x >= 1, we know e-x > e-x3/2 over the interval of integration, hence the given integral converges if the integral of e-x converges. But the integral of e-x from 1 to infinity does converge; it is equal to limR -> infinity -e-x|1R = -0 - -e-1 = 1/e.

• Use a comparison to determine whether the following improper integral converges or diverges; completely justify your answer:

(x2 + x)/(x3 + 2) dx
1

Whether the integral converges or not depends on what hapens when x is very large. But for large x, x is negligible compared to x2, so (x2 + x) is about the same as x2. Likewise, (x3 + 2) is about the same as x3. Thus (x2 + x)/(x3 + 2) is about the same as x2/x3 = 1/x, and we know the integral of 1/x from 1 to infinity diverges. But that was just a heuristic argument. Here's the rigorous argument using the comparison theorem. Since we now expect that we want to show divergence, we need to find an f(x) which is less than (x2 + x)/(x3 + 2) but for which the integral diverges. Note that (x2 + x)/(x3 + 2) > x2/(x3 + 2) > x2/(x3 + 2x3) = x2/(3x3) = (1/3)(1/x). But of course the integral from 1 to infinity of (1/3)(1/x) diverges, since it does for 1/x (one third of infinity is still infinity!).