momin (rrm497) – Homework 04 – cheng – (58520)
1
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printout
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15
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001
10.0 points
Find the area between the graph of
f
and
the
x
axis on the interval [0
,
4] when
f
(
x
) = 2
x
−
x
2
.
1.
Area = 7 sq.units
2.
Area = 5 sq.units
3.
Area = 4 sq.units
4.
Area = 8 sq.units
correct
5.
Area = 6 sq.units
Explanation:
The graph of
f
is a parabola opening down
wards and crossing the
x
axis at
x
= 0 and
x
= 2.
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
2
0
(2
x
−
x
2
)
dx
−
integraldisplay
4
2
(2
x
−
x
2
)
dx .
Now
integraldisplay
2
0
(2
x
−
x
2
)
dx
=
bracketleftBig
x
2
−
1
3
x
3
bracketrightBig
2
0
=
4
3
,
while
integraldisplay
4
2
(2
x
−
x
2
)
dx
=
bracketleftBig
x
2
−
1
3
x
3
bracketrightBig
4
2
=
−
20
3
.
Consequently,
Area = 8 sq.units
.
keywords: integral, graph, area
002
10.0 points
Find the area enclosed by the graphs of
f
(
x
) = cos
x ,
g
(
x
) = sin
x
on [0
, π
].
1.
area = 2
√
2
correct
2.
area = 4(
√
2 + 1)
3.
area = 4
√
2
4.
area =
√
2 + 1
5.
area = 2(
√
2 + 1)
6.
area =
√
2
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
=
integraldisplay
π
0

cos
x
−
sin
x

dx .
But, as the graphs
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momin (rrm497) – Homework 04 – cheng – (58520)
2
y
θ
π/
2
π
cos
θ
:
sin
θ
:
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
=
integraldisplay
π/
4
0
{
cos
θ
−
sin
θ
}
dθ
−
integraldisplay
π
π/
4
{
cos
θ
−
sin
θ
}
dθ
=
A
1
−
A
2
.
But by the Fundamental Theorem of Calcu
lus,
A
1
=
bracketleftBig
sin
θ
+ cos
θ
bracketrightBig
π/
4
0
=
√
2
−
1
,
while
A
2
=
bracketleftBig
sin
θ
+ cos
θ
bracketrightBig
π
π/
4
=
−
(1 +
√
2)
.
Consequently,
area =
A
1
−
A
2
= 2
√
2
.
003
10.0 points
Compute the area between the graphs of
f
(
x
) = 8 sin 2
x
and
g
(
x
) = 6 sin
x
+ 2 sin 2
x
on [0
, π/
2].
1.
Area =
5
2
sq.units
2.
Area =
13
4
sq.units
3.
Area =
11
4
sq.units
4.
Area =
9
4
sq.units
5.
Area = 3 sq.units
correct
Explanation:
The required area is given by
I
=
integraldisplay
π/
2
0

f
(
x
)
−
g
(
x
)

dx.
Now the graphs of
f, g
intersect when
8 sin 2
x
= 6 sin
x
+ 2 sin 2
x ,
i.e.
, when
6(sin 2
x
−
sin
x
)
= 6 sin
x
(2 cos
x
−
1) = 0
,
since sin 2
x
= 2 sin
x
cos
x
. Thus the points of
intersection are
x
= 0
, π/
3.
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 Fall '08
 RAdin
 Calculus, Cone, dx

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