MTH 1126 - Test #1 - Solutions
Spring 2008
Pat Rossi
Name
Instructions. Show CLEARLY how you arrive at your answers.
1.
R
√
(4x3 + 3x2 − 6x + 3 x + sec2 (x)) dx
=4
i.e.,
h
x4
4
R
i
+3
h
x3
3
i
−6
h
x2
2
i
+3
∙
3
x2
( 32 )
¸
=
|{z}
- re-write %
R³
1
´
4x3 + 3x2 − 6x + 3x 2 + sec2 (x) dx
3
+ tan (x) + C = x4 + x3 − 3x2 + 2x 2 + tan (x) + C
√
3
(4x3 + 3x2 − 6x + 3 x + sec2 (x)) dx = x4 + x3 − 3x2 + 2x 2 + tan (x) + C
2. Use the “f − g” method to compute the area bounded by the graphs of f (x) = x3 and
g (x) = 4x.
First, graph the functions and find the points of intersection.
To find the points of intersection, set f (x) = g (x) .
f (x) = x3 = 4x = g (x)
⇒ x3 = 4x
⇒ x3 − 4x = 0
⇒ x (x2 − 4)
⇒ x (x + 2) (x − 2) = 0
⇒ x = 0, x = −2; x = 2
The points of intersection are: (−2, −8) , (0, 0) , and (2, 8) .
(2, 8)
g(x) = 4x
f(x) = x3
(0, 0)
(!2, !8)
Since 4x ≥ x3 over the interval [0, 2] , the area of the bounded region is given by
R2
0
h
(4x − x3 ) dx = 2x2 − 14 x4
i2
0
³
´
³
´
= 2 (2)2 − 14 (2)4 − 2 (0)2 − 14 (0)4 = 4
Area = 4
R
8
3. Suppose that
2 (f (x) + g (x)) dx = 10;
R8
Compute 4 f (x) dx.
First, we need to find the value of
R8
Since
R8
2
2
(f (x) + g (x)) dx =
f (x) dx =
i.e.,
R8
2
R8
2
i.e.,
4
4
R2
f (x) dx =
f (x) dx = 9
4
f (x) dx +
R2
4
R8
f (x) dx +
2
g (x) dx = 5; and that
2
f (x) dx.
f (x) dx +
f (x) dx = 5
R8
R8
2
2
(f (x) + g (x)) dx −
Next observe that
Thus,
R8
R8
R8
R8
2
R8
2
g (x) dx, it follows that
g (x) dx = 10 − 5 = 5
f (x) dx =
R8
2
Z 8
|4
f (x) dx
{z
}
this is what we want
f (x) dx = 4 + 5 = 9
2
R2
4
f (x) dx = 4.
4. Compute:
R x=1
3
(2x3 + 3) x2 dx =
x=0
1. Is u-sub appropriate?
a. Is there a composite function?
3
Yes. (2x3 + 3)
Let u = 2x3 + 3
b. Is there an “approximate function/derivative pair”?
Yes. (2x3 + 3) → x2
Let u = 2x3 + 3
2. Compute du
2x3 + 3
6x2
6x2 dx
x2 dx
u
=
du
=
dx
du =
1
du =
6
⇒
⇒
⇒
When x = 0, u = 2x3 + 3 = 3
When x = 1, u = 2x3 + 3 = 5
3. Analyze in terms of u and du.
R x=1 ³
x=0
´3
2
2x3 + 3 x
| {zdx}=
|
{z
}
u3
1
du
6
R u=5 3 1
u du
u=3
6
=
1 R u=5
6 u=3
u3 du
4. Integrate in terms of u
1 R u=5
6 u=3
i.e.,
u3 du =
R x=1
x=0
1
6
h
iu=5
u4
4 u=3
3
=
1
24
(2x3 + 3) x2 dx =
(54 − 34 ) =
68
3
68
3
5. Find the area bounded by the graphs of f (x) = 1 − x2 and g (x) = x2 − 1. (Partition
the proper interval, build the Riemann Sum, derive the appropriate integral.)
1. First, graph the functions and find the points of intersection.
To find the points of intersection, set f (x) = g (x) .
f (x) = 1 − x2 = x2 − 1 = g (x)
⇒ 1 − x2 = x2 − 1
⇒ 2 − 2x2 = 0
⇒ 1 − x2 = 0
⇒ (x + 1) (x − 1) = 0
⇒ x = −1; x = 1
3
Points of intersection are (−1, 0) and (1, 0) .
f(x) = 1 ! x2
(1, 0)
(!1, 0)
g(x) = x2 ! 1
2. Partition the interval spanned by the region into sub-intervals of length ∆x.
3. Above the ith subinterval, inscribe a rectangle of width ∆x.
f(x) = 1 ! x2
(xi , 1! xi2)
(1, 0)
(!1, 0)
xi
(1! xi2) ! (xi2 !1)
(xi , xi2 !1)
g(x) = x ! 1
2
)x
Height of the ith rectangle = [(1 − x2i ) − (x2i − 1)] = 2 − 2x2i
Width of the ith rectangle = ∆x
Area of the ith rectangle = (2 − 2x2i ) ∆x
4. Approximate the area of the region by adding up the areas of the rectangles.
Area ≈
Pn
i=1
(2 − 2x2i ) ∆x
5. Let ∆x → 0.
Area = lim∆x→0
³
´
Pn
2
i=1 (2 − 2xi ) ∆x =
³
´
R3
2 (1) − 23 (1)3 − 2 (−1) − 23 (−1)3 =
Bounded area =
8
3
4
h
2 3
2
−1 (2 − 2x ) dx = 2x − 3 x
8
3
i1
−1
=
R
6. Compute:
tan (3x) sec2 (3x) dx =
1. Is u-sub appropriate?
a. Is there a composite function?
Yes. tan (3x) also sec (3x) also sec2 (3x)
Let u = 3x or u = sec (3x)
b. Is there an “approximate function/derivative pair”?
Yes. tan (3x)→sec2 (3x)
|
{z
function
}
|
{z
deriv
Let u = tan (3x)
}
Remark 1 Since criteria a and b yield different choices for u, we will go
with criterin b, the “approximate function/derivative pair,” and we will let
u = tan (3x)
2. Compute du
⇒
⇒
⇒
u =
=
du =
1
du =
3
du
dx
tan (3x)
3 sec2 (3x)
3 sec2 (3x) dx
sec2 (3x) dx
3. Analyze in terms of u and du.
R
R
tan (3x)sec2 (3x) dx= = u 13 du =
|
{z
u
}|
{z
1
du
3
}
1
3
R
u du
4. Integrate in terms of u
1
3
R
u du =
1 u2
3 2
+ C = 16 u2 + C
5. Re-write in terms of x
R
1
tan (3x) sec2 (3x) dx = tan2 (3x) + C
|6
{z
}
1 2
u +C
6
5
7. The graph of f (x) is shown below. Compute
R3
−3
f (x) dx
(0, 3)
(!3, 0)
(3, 0)
(0, 0)
R
3
Observe that between x = −3 and x = 3, f (x) ≥ 0. Therefore, −3
f (x) dx is equal to
the area bounded by the graph of f (x) and the x-axis, between x = −3 and x = 3.
Thus,
R3
³
−3 f (x) dx = (area of triangle with base and height = 3)+ area of
= 12 b · h + 14 πr2 = 12 (3) (3) + 14 π (3)2 =
9
2
6
+
9π
4
=
18+9π
4
1
4
´
circle of radius 3