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.