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Final Exam - Solutions
- (7 pts) Compute (^) dxd {x^2 esin(x)}. Please do not simplify your answer. Solution: 2 xesin(x)^ + x^2 esin(x)^ cos(x)
- (8 pts) Suppose that f is a continuous function and set F (x) =
∫ (^) x 2 0 f^ (y) dy. Compute F ′(x). You will receive more partial credit for an incorrect answer if you define intermediate functions and clearly indicate the rules that you are using. Solution: Set G(x) =
∫ (^) x 0 f^ (y) dy, so^ G
′(x) = f (x), and
F ′(x) =
d dx
G(x^2 ) = G′(x^2 )2x = 2xf (x^2 ).
- (8 pts) Let f be a differentiable function whose domain is [− 2 , 2]. Further suppose that f (0) = 0 and f ′^ is given by the following graph. Again, you are given the graph of f ′^ not the graph of f. Please answer the following questions using interval notation. For example [a, b) ∪ {c, d} ∪ (e, f ]. Don’t worry to much about getting the numbers exactly right. If you are within 1/4 of the true values, you will receive full credit.
x -2 -1 1 2
f ′(x)
(a) (2 pts) At which of the points in the interval (− 2 , 2) is f increasing? Solution: (− 2 , 3 /2).
(b) (2 pts) At which of the points in the interval (− 2 , 2) is f concave up? Solution: The function f is concave up when f ′′^ > 0. These are the points where f ′^ is increasing: (− 3 / 2 , 0). (c) (2 pts) Is f (−1) positive or negative? Solution: It is negative. (d) (2 pts) Does f have a global max? If so, where does it occur? Solution: The function f has global max at x = 3/2 by the first derivative test.
- (7 pts) Compute (^) dxd {cos(x) tan-1(x^2 )}. Please do not try to simplify your answer. The function tan-1^ is the inverse of tan not 1/ tan. Solution: − sin(x) tan-1(x^2 ) + 2 x1+^ cos(x 4 x ).
- (9 pts) Let f be function with domain [0, 2] and f (x) = x^3 − 6 x^2 + 9x + 1 when 0 ≤ x ≤ 2.
(a) (3 pts) Find the critical points of the function f that lie in the interval (0, 2). Solution: We see that f ′(x) = 3x^2 − 12 x + 9 = 2(x − 1)(x − 3), so the only critical point in the interval is 1. (b) (3 pts) What is the (global) maximum value of f on the interval [0, 2] and where does it occur? Put a circle around the maximum value and put a box around the number in the domain where the maximum occurs. Solution: We just check the critical points and the endpoints: f (0) = 1, f (1) = 5, f (2) = 3. So the maximum value is 5 and it occurs at 1. (c) (3 pts) What is the (global) minimum value of f on the interval [0, 2] and where does it occur? Put a circle around the minimum value and put a box around the number in the domain where the minimum is occurs. Solution: We have already looked at the critical points, so it is clear that the minimum value is 1 and it occurs at 0.
- (6 pts) Compute (^) dxd {ln(x^3 )}. Solution: (^) dxd {ln(x^3 )} = 3/x.
- (6 pts) Let f be a differentiable and invertible function, set g(x) = f −^1 (x), and
- (7 pts) Compute
∫ (^) t
0
x cos(x^2 ) dx. Solution: We make the substitution u = x^2 , so ∫ (^) t
0
x cos(x^2 ) dx =
∫ (^) t 2
0
cos(u) 2
du =
[
sin(u)/ 2
]u=t^2 u=
= sin(t^2 )/ 2.
Set θ = sin−^1 x, so cos θ =
1 − x^2 and tan θ = sin θ/ cos θ = x/
1 − x^2.
- (7 pts) Compute
∫ (^) t
0
x sin(x) dx. Solution: We use integration by parts with F (x) = x, f (x) = 1, G(x) = − cos(x) and g(x) = sin(x), so ∫ (^) t
0
x sin(x) dx =
[
−x cos(x)
]x=t x=0 +
∫ (^) t
0
cos(x) dx
=
[
sin(x) − x cos(x)
]x=t x=0 = sin(t)^ −^ t^ cos(t).
- (7 pts) Let f (x) = x^3 + 2x^2 and let g(x) = 5 x + 6. Setup the integral that you would use to compute the area of the bounded re- gion that is contained between f and g. You do not need to actually compute this inte- gral, but your final answer should not con- tain “f ” or “g”. It should also not contain any absolute value signs. The following plot is provided for you convenience.
x -4 -3 -2 -1 1 2 3 4
f(x) g(x)
Solution: The area is given by: ∫ (^) − 1
− 3
f (x) − g(x) dx +
− 1
g(x) − f (x) dx
− 3
x^3 + 2x^2 − 5 x − 6 dx +
− 1
−x^3 − 2 x^2 + 5x + 6 dx
- (7 pts) Let S be a sphere with radius r. Write down the integral that you would use to compute the vol- ume of the bottom fourth of the sphere. Please do not compute the integral.
Solution: Let h = 0 correspond to the bottom of the sphere, and let f (h) denote the radius of the circular cross section of the sphere which is at the height h. Then f 2 (h) + (r − h)^2 = r^2 , so f 2 (h) = 2h(r − h). This means that the area of the cross section at height h is π 2 hr − h^2 , and the desired integral is:
π
∫ (^) r/ 2
0
2 hr − h^2 dh.
- (7 pts) Compute
∫ (^) t
0
3 x^5 sin(x^3 ) dx. You will receive more credit for an incorrect answer if you clearly indicate what you are doing. You do not need to simplify your answer. Solution: We make the substitution u = x^3 followed by integration by parts with with F (u) = u and G(u) = − cos(u) to see that ∫ (^) t
0
x^5 sin(x^3 ) dx =
∫ (^) t 3
0
u sin(u) dx
[
−u cos(u)
]u=t 3 u=
∫ (^) t 3
0
cos(u) du
[
sin(u) − u cos(u)
]u=t^3 u= = sin(t^3 ) − t^3 cos(t^3 ).
- (7 pts) Compute
∫ (^) t 0 cos
(^3) (x) dx. You will receive more credit for an incorrect answer if you clearly indicate what you are doing. You do not need to simplify your answer. Solution: Making the change of variable u = sin(x) gives ∫ (^) t
0
cos^3 (x) dx =
∫ (^) t
0
{ 1 − sin^2 (x)} cos(x) dx
∫ (^) sin(t)
sin(0)
1 − u^2 du
[
u − u^3 / 3
]sin(t) sin(0) = sin(t) − sin^3 (t)/ 3.
