Download - Calculas 2 practice exam
Math 2100 (Section 03) Fall 2014 Practice Exam 1 -Solutions
(1) Compute
d
dxex
3−5
.
Solution. Recall ddx e
g(x) = g′(x). Put g(x) = x3− 5, and ddx e
x3−5 =ddx e
g(x) = eg(x)g′(x) = ex3−5(3x2). �
(2) Compute
d
dxln (6x2 − 3x+ 1)
.
Solution. Recall ddx ln (g(x)) = g′(x)
g(x) . Put g(x) = 6x2−3x+1. Then
ddx ln (6x2 − 3x+ 1) = g′(x)
g(x) = 12x−36x2−3x+1
. �
(3) Evaluate the integral ∫ √x+
1
xdx.
Solution. =∫x1/2 + x−1 dx = 2
3x3/2 + ln |x|+ C. �
(4) Evaluate the integral ∫ 0
−13e−t + t dt.
Solution. = 3 · e−t
−1 + t2
2 |−10 , i.e. ((3 e
−1 + 12)− (3 e0
−1 + 0)) = −3e+ 12 +
3. �
(5) A rock is dropped from the top of a 500-foot cliff. Its velocity attime t seconds is v(t) = −32t feet per second. Find the displacementof the rock during the time interval 2 ≤ t ≤ 4.
Proof. The displacement is given by∫ 42 v(t) dt =
∫ 42 −32t dt = −32t2
2 |42 =
−16t2|42 = (−256 + 64) = −192. (That is, the rock fell by 192 feetduring the time interval 2 ≤ t ≤ 4.) �
(6) Find the area of the following shaded region:
Solution. The area is given by∫ 31 x+ 1
x dx =∫ 31 x+x−1 dx =
(x2
2 + ln |x|)|31 =
(92 + ln 3− 12 − 0) = 4 + ln 3. �
(7) Find the area of the region bounded by y = x2 − 1 and y = 3.
Solution. Plot the two curves first:1
2
Identify the intersection points first: set x2− 1 = 3⇒ x2 = 4, i.e.x = ±2. So the area of the bounded region is∫ 2
−23− (x2 − 1) dx =
∫ 2
−24− x2 dx
= 4x− x3
3|2−2
= (8− 8
3)− (−8 +
8
3) = 16− 16
3=
32
3.
�
(8) Suppose that money is deposited steadily in a savings account so that$16,000 is deposited each year. Determine the balance at the end of4 years if the account pay 8% interest compounded continuously.
Solution. In the formula∫ N0 Ker(N−t) dt, we have N = 4, K = 16000
and r = 0.08. So the balance at the end of 4 years is∫ 4
0= 16000e0.08(4−t) dt = 16000
∫ 4
0e0.32e−0.08t dt
= 16000e0.32∫ 4
0e−0.08t dt
= 16000e0.32(e−0.08t−0.08|40
)= 16000e0.32
(e−0.32−0.08− 1
−0.08
)≈ 75426.
So the balance at the end of 4 years will be roughly $75426. �
The following formulas will be provided on Exam 1:
3
• Consumers’ surplus: ∫ A
0f(x)−B dx.
• The future value of a continuous income stream:∫ N
0Ker(N−t) dt.
• The volume of a solid of revolution:∫ b
0π(g(x))2 dx.