Calculus II Quiz #4 March 6, 2002 Name____________________ R.  Hammack Score ______

(1) Evaluate the definite integrals. Simplify your answer as much as possible
(a)  ==

ux
dudx

(b)  =ln(e+5) - ln(6) = ln((e+5)/6)

u =
du = dx

(2)
Find the derivative of the function  F(x) =
Using the chain rule and Part 2 of the Fundamental Theorem of Calculus, we get
F '(x) = sin( -(x+ln(x)) )(1 + 1/x)

(3) A freight train, moving with constant acceleration, travels 20 miles in half an hour.  At the beginning of the half-hour period, it has a velocity of 10 miles per hour.  What is its velocity at the end of the half-hour period?

Let's set the clock so that t=0 represents the beginning of the half-hour period and t = 1/2 represents the end.
Represent the constant acceleration as a, so the velocity is v(t) = = at +C.
Since it's given that 10 = v(0) = a(0) + C, we get C = 10.
Therefore the velocity is v(t) = at +10. If we can just find a, the answer to the question will be v(1/2).

To use the information about position (i.e. the fact that the train went 20 miles), we must now construct the position function.

Notice that s(t) = = = ,
and from 0 = s(0) = we obtain K=0.
Therefore, position is s(t) =

Since the train went 20 miles in half an hour, we know that 20 = s(1/2) =
which becomes 20 = a/8 +5. Solving for a, we get a = 120.
Therefore, we can finally write velocity as v(t) = 120t +10.

Here is the answer to the question. The velocity at the end of the half-hour period is
v(1/2) = 120(1/2)+10 = 70 miles per hour.