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QUIZ ON CHAPTER 6
SOLUTIONS
MATH 150 – SPRING 2011 – KUNIYUKI
105 POINTS TOTAL, BUT 100 POINTS = 100%
Show all work, simplify as appropriate, and use “good form and procedure” (as in class).
Box in your final answers!
No notes or books allowed. A scientific calculator is allowed.
Assume that all graphs here are in the usual
xy
plane.
Distances and lengths are measured in meters throughout this exam.
Write appropriate units in your final answers.
You may not use absolute value (or similar short cuts) in your final integral setups.
Note
: The functions here are continuous on the intervals of interest. This guarantees integrability
and the validity of the Test Value Method to determine relative positions on an interval (i.e.,
which graph is on top vs. on bottom, or on the right vs. on the left). The FTC applies.
1)
Find the area of the region
R
bounded by the graphs of
y
=
2
x
2
+
3
x
±
4
and
y
=
x
2
+
5
x
+
4
.
Identify any intersection points. You do not
have to sketch
the region or find intercepts. Evaluate
your integral completely. Show all
work. (27 points)
In both of the given equations,
y
is already expressed in terms of
x
.
We hope to do a “
dx
scan.”
Find the
x
coordinates of the intersection points of the corresponding graphs by solving
the following system for
x
; they will be the limits of integration.
y
=
2
x
2
+
3
x
±
4
y
=
x
2
+
5
x
+
4
²
³
´
µ
´
¶
2
x
2
+
3
x
±
4
=
x
2
+
5
x
+
4
x
2
±
2
x
±
8
=
0
x
±
4
()
x
+
2
=
0
x
±
4
=
0
x
=
4
or
x
+
2
=
0
x
=
±
2
Our limits of integration will apparently be
a
=
±
2
and
b
=
4
.
We will use the equation
y
=
x
2
+
5
x
+
4
to determine the
y
coordinates of the
intersection points. (We could also use the equation
y
=
2
x
2
+
3
x
±
4 .)
x
=
±
2
²
y
=
±
2
2
+
5
±
2
+
4
²
y
=
±
2
²
Point
±
2,
±
2
x
=
4
²
y
=
4
2
+
54
+
4
²
y
=
40
²
Point 4, 40
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View Full Document The graphs of
y
=
x
2
+
5
x
+
4 and
y
=
2
x
2
+
3
x
±
4 are parabolas.
Which parabola forms the top boundary of
R
, and which forms the bottom boundary?
We can test an
x
value in the interval
±
2, 4
()
, say
x
=
0 .
(Then, we’re identifying
y
intercepts, also.)
Graph of
y
=
2
x
2
+
3
x
±
4 :
y
=
20
2
+
30
±
4
²
y
=
±
4
²
y
int. is
0,
±
4
³
´
µ
¶
Graph of
y
=
x
2
+
5
x
+
4:
y
=
0
2
+
50
+
4
±
y
=
4
±
y
int. is
0, 4
²
³
´
µ
4
>
±
4 , so this graph forms the top boundary of
R
.
Here is a sketch of
R
; the
x
 and
y
axes are scaled differently:
The region
R
is wellsuited to a “
dx
scan.”
We’re not revolving
R
about any axis, so the fact that the
x
axis and the
y
axis pass
through the interior of
R
does not pose a problem.
Area
A
=
x
2
+
5
x
+
4
()
y
top
x
±²
³
³´
³³
±
2
x
2
+
3
x
±
4
y
bottom
x
³
²
³
´
´
´
µ
¶
·
·
·
dx
±
2
4
¸
=
x
2
+
5
x
+
4
±
2
x
2
±
3
x
+
4
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This note was uploaded on 09/08/2011 for the course MATH 150 taught by Professor Bart during the Spring '06 term at Mesa CC.
 Spring '06
 Bart
 Math, Calculus

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