MATH 106, Section D
Test # 1
SOLUTIONS
1. (20 points) Evaluate the integral.
cos(ln x)
x
dx
u= lnx=⇒du =
1
x
dx =⇒
cos(ln x)
x
dx =cos udu = sin u+C= sin ln x+C.
2. (20 points) Find the minimal value of nsuch that R
n
approximates the value of the integral with the absolute error
less than or equal to 0.007.
e
1
xln xdx
First of all, we want to find K
1
, to do this we have to find critical points of f
(x)and compare the values of f
at them
with the values of f
at the end-points of the interval. Among all these values we need the one with the largest absolute
value, this absolute value is K
1
, i.e., the minimal number that bounds |f
(x)|from above. Hence,
f(x) = xln x=⇒f
(x) = ln x+ 1 =⇒f
(x) =
1
x
= 0 =⇒no critical points.
Now we have the following values: f
(1) = 1,f
(e) = 2. Thus K
1
= 2, therefore, the bound gives us
0.007 ≤K
1
(b−a)
2
2n=2 (e−1)
2
2n=⇒n≤(e−1)
2
0.007 ≈421.7846 =⇒n= 422 .
3. (20 points) Find the area of the region in the xy-plane bounded by the curves y=e
2x
,y=e
x
, and x= ln2.
A=
ln 2
0
e
2x
−e
x
dx =
ln 2
0
e
2x
dx −
ln 2
0
e
x
dx =
1
2
e
2x
ln 2
0
−e
x
ln 2
0
=
1
2
e
2 ln 2
−e
0
−e
ln 2
−e
0
=
1
2
e
ln 2
2
−1−(2 −1) =
1
2
(4 −1) −1 =
3
2
−1 =
1
2
.
4. (20 points) Find the length of the curve. y= lnx−
x
2
8
, x ∈[1,4]
L=
4
1
1 + ln x−
x
2
8
2
dx =
4
1
1 +
1
x
−
x
4
2
dx =
4
1
1 +
1
x
2
−2·
1
x
·
x
4
+
x
4
2
dx
=
4
1
1 +
1
x
2
−
1
2
+
x
4
2
dx =
4
1
1
x
2
+
1
2
+
x
4
2
dx =
4
1
1
x
+
x
4
2
dx =
4
1
1
x
+
x
4
dx
=
4
1
1
x
+
x
4
dx =
4
1
1
x
dx +
1
4
4
1
xdx = ln|x|
4
1
+
1
8
x
2
4
1
= ln4 −ln1 +
1
8
(16 −1) = 2 ln 2 +
15
8
.
5. (20 points) Find the volume of the solid of revolution formed when the region bounded by y= 1−√x,y=x+ 1,
and x= 1 is revolved about the line y= 3.
The area of the cross section: A(x) = π(3 −(1 −√x))
2
−π(3 −(x+ 1))
2
=π(2 + √x)
2
−π(2 −x)
2
=π4 + 4√x+x−4−4x+x
2
=π4√x+ 5x−x
2
.
V=
1
0
A(x)dx =π
1
0
4√x+ 5x−x
2
dx =π
4
3/2
x
3/2
1
0
+
5
2
x
2
1
0
−
1
3
x
3
1
0
=π
8
3
+
5
2
−
1
3
=π
16+15−2
6
=π
29
6
=
29π
6
.
1
