© Scarborough Math 131 ... (20pts) NAME (printed neatly): ________________________________

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© Scarborough
Math 131
Fall 2007
Quiz 6 KEY
(20pts) NAME (printed neatly): ________________________________
(10pts) Section Number (circle correct section): 502 (10:20am)
503 (11:30am)
506 (4:10pm)
From September 1 to September 17, the rate of change in temperature is
t (n) = −0.12n 2 + 1.92n − 4.68 maximum degrees F per September day where n is
the day in September, 1 ≤ n ≤ 17 .
a. (5pts) What are the input units of t(n)?
day in September or d day in September or days or d days
b. (5pts) What are the output units of t(n)?
maximum degrees F per day or maximum degrees F per day in September
c. (5pts) What are the units of the area of the region between the input axis
and the output axis?
(maximum degrees F per Sept day)(day in Sept) = maximum degrees F
d. (10pts) Find the points A and B where the graph crosses the x-axis, and
let a be the x-coordinate of point A and let b be the x-coordinate of point B
such that a < b.
Y1 = t(n)
Graph and find zeros.
A(3, 0) so a = 3
B(13, 0) so b = 13
e. (5pts) What does the area of the region lying above the axis represent?
an increase in temperature
© Scarborough
Math 131
Fall 2007
Quiz 6 KEY
2
f. (10pts) Using rectangles of width 2, find the approximate area of the graph
below the x-axis. Use appropriate integration symbols. [Recall, since I did
not specify, you are to use midpoints.]
3
17
− ( ∫ t (n)dn + ∫ t (n)dn) ≈ −([t (2)](2) + [t (14) + t (16)](2)) = −(−1.32 + −1.32 + −4.68)(2) = 14.64
1
13
is the area of the decrease in maximum degrees F from Sept 1 to 3 and from
Sept 13 to 17.
g. (10pts) Using rectangles of width 2, find the approximate area of the graph
above the x-axis. Use appropriate integration symbols.
13
∫ t (n)dn ≈ [t (4) + t (6) + t (8) + t (10) + t (12)](2) = (1.08 + 2.52 +3 + 2.52 + 1.08)(2)
3
= (10.2)(2) = 20.4 is the area of increase in maximum degrees F from Sept 3 to
Sept 13.
a
h. (10pts) Estimate ∫ t (n)dn using rectangles of width 2. See part d for value
1
of a.
3
∫ t (n)dn ≈ (t (2))(2) = (−1.32)(2) = −2.64
is the decrease in maximum degrees F
1
from Sept 1 to 3.
17
i.
(10pts) Estimate
∫ t (n)dn
using your calculations in the above parts.
1
17
∫ t (n)dn ≈ 20.4 − 14.64 = 5.76 is the increase in maximum degrees F from Sept 1 to
1
Sept 17.
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