Calculus II |
Quiz #1
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February 15, 2002
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Name____________________ |
R. Hammack
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Score ______
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(1) Find the following antiderivatives.
(a) ![[Graphics:Images/quiz1solved_gr_1.gif]](Images/quiz1solved_gr_1.gif)
![[Graphics:Images/quiz1solved_gr_2.gif]](Images/quiz1solved_gr_2.gif)
(b) ![[Graphics:Images/quiz1solved_gr_3.gif]](Images/quiz1solved_gr_3.gif)
![[Graphics:Images/quiz1solved_gr_4.gif]](Images/quiz1solved_gr_4.gif)
(c) ![[Graphics:Images/quiz1solved_gr_5.gif]](Images/quiz1solved_gr_5.gif)
(d) ![[Graphics:Images/quiz1solved_gr_6.gif]](Images/quiz1solved_gr_6.gif)
![[Graphics:Images/quiz1solved_gr_7.gif]](Images/quiz1solved_gr_7.gif)
![[Graphics:Images/quiz1solved_gr_8.gif]](Images/quiz1solved_gr_8.gif)
![[Graphics:Images/quiz1solved_gr_9.gif]](Images/quiz1solved_gr_9.gif)
(e) ![[Graphics:Images/quiz1solved_gr_10.gif]](Images/quiz1solved_gr_10.gif)
![[Graphics:Images/quiz1solved_gr_11.gif]](Images/quiz1solved_gr_11.gif)
![[Graphics:Images/quiz1solved_gr_12.gif]](Images/quiz1solved_gr_12.gif)
![[Graphics:Images/quiz1solved_gr_13.gif]](Images/quiz1solved_gr_13.gif)
(2) The graph of a function f(x) has slope
at
each point (x, f(x)).
Also, the graph of f(x) passes through the point (1,
5). Find the function f(x).
The information says slope = f '(x) =
.
This means f(x) =![[Graphics:Images/quiz1solved_gr_16.gif]](Images/quiz1solved_gr_16.gif)
.
Thus f(x) = ![[Graphics:Images/quiz1solved_gr_18.gif]](Images/quiz1solved_gr_18.gif)
To determine f completely, we just need to find C. To do this, use
the fact that f(1)=5.
5=f(1)= ![[Graphics:Images/quiz1solved_gr_19.gif]](Images/quiz1solved_gr_19.gif)
5= 0 + 3/4 +C
Therefore C = 17/4, and the answer to the question is f(x)
= ![[Graphics:Images/quiz1solved_gr_20.gif]](Images/quiz1solved_gr_20.gif)