Hw05 - badruddin(ssb776 HW05 Radin(57410 This print-out...

badruddin (ssb776) – HW05 – Radin – (57410) 1
This print-out should have 19 questions.
Multiple-choice questions may continue on
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before answering.
001 10.0 points
For which one of the following shaded re-
gions is its area represented by the integral
integraldisplay
4
0
braceleftbigg
1
4
x + ( x + 1)
bracerightbigg
dx ?
1.
2 4 6 − 2
2
− 2
− 4
− 6
correct
2.
2 4 6 − 2
2
− 2
− 4
− 6
3.
2 4 6 − 2
2
4
6
4.
2 4 6 − 2
2
4
6
− 2
5.
2 4 6 − 2
2
4
6
− 2
Explanation:
If f ( x ) ≥ g ( x ) for all x in an interval [ a, b ],
then the area between the graphs of f and g
is given by
Area =
integraldisplay
b
a
braceleftBig
f ( x ) − g ( x )
bracerightBig
dx .
When
f ( x ) =
1
4
x, g ( x ) = − ( x + 1) ,
therefore, the value of
integraldisplay
4
0
braceleftBig
1
4
x + ( x + 1)
bracerightBig
dx
is the area of the shaded region
2 4 6 − 2
2
− 2
− 4
− 6
.
002 10.0 points

badruddin (ssb776) – HW05 – Radin – (57410) 2
Find the area between the graph of f and
the x -axis on the interval [0 , 6] when
f ( x ) = 3 x − x
2
.
1. Area = 30 sq.units
2. Area = 27 sq.units correct
3. Area = 29 sq.units
4. Area = 31 sq.units
5. Area = 28 sq.units
Explanation:
The graph of f is a parabola opening down-
wards and crossing the x -axis at x = 0 and
x = 3. Thus the required area is similar to
the shaded region in the figure below.
graph of f
In terms of definite integrals, therefore, the
required area is given by
integraldisplay
3
0
(3 x − x
2
) dx −
integraldisplay
6
3
(3 x − x
2
) dx .
Now
integraldisplay
3
0
(3 x − x
2
) dx =
bracketleftBig
3
2
x
2
−
1
3
x
3
bracketrightBig
3
0
=
9
2
,
while
integraldisplay
6
3
(3 x − x
2
) dx =
bracketleftBig
3
2
x
2
−
1
3
x
3
bracketrightBig
6
3
= −
45
2
.
Consequently,
Area = 27 sq.units .
keywords: integral, graph, area
003 10.0 points
Find the area enclosed by the graphs of
f ( x ) =
1
2
cos x , g ( x ) =
1
2
sin x
on [0 , π ].
1. area =
√
2 correct
2. area = 4
√
2
3. area = 2
√
2
4. area = 2(
√
2 + 1)
5. area =
√
2 + 1
6. area = 4(
√
2 + 1)
Explanation:
The area between the graphs of y = f ( x )
and y = g ( x ) on the interval [ a, b ] is expressed
by the integral
A =
integraldisplay
b
a
| f ( x ) − g ( x ) | dx ,
which for the given functions is the integral
A =
1
2
integraldisplay
π
0
| cos x − sin x | dx .
But, as the graphs
y
θ
π/ 2
π
cos θ :
sin θ :

badruddin (ssb776) – HW05 – Radin – (57410) 3
of y = cos x and y = sin x on [0 , π ] show,
cos θ − sin θ
braceleftBigg
≥ 0 , on [0 , π/ 4],
≤ 0 , on [ π/ 4 , π ].
Thus
A =
1
2
integraldisplay
π/ 4
0
{ cos θ − sin θ } dθ
−
1
2
integraldisplay
π
π/ 4
{ cos θ − sin θ } dθ = A
1
− A
2
.
But by the Fundamental Theorem of Calcu-
lus,
A
1
=
1
2
bracketleftBig
sin θ + cos θ
bracketrightBig
π/ 4
0
=
1
2
(
√
2 − 1) ,
while
A
2
=
1
2
bracketleftBig
sin θ + cos θ
bracketrightBig
π
π/ 4
= −
1
2
(1 +
√
2) .
Consequently,
area = A
1
− A
2
=
√
2 .
004 10.0 points
Find the area of the shaded region in
x
y
bounded by the graphs of
f ( x ) = x
3
− x
2
− 5 x, g ( x ) = x .