Math 1172 (Buenger)
Solutions to Section 6.2
Solution: Let u = ln |cos(7x)|. Then
U-sub review:
1.
Z
du =
2x sin(x2 )dx
and
Solution: Let u = x . Then du = 2x dx
. We have the following suband dx = du
2x
stitution:
Z
Z
du
2
2x sin(x )dx =
2x sin(u)
2x
Z
=
sin(u) du
du
.
−7 tan(7x)
Thus
Z
tan(7x) ln |cos(7x)| dx
Z
du
tan(7x)u
−7 tan(7x)
Z
−1
=
u du
7
−1 u2
·
+C
=
7
2
−(ln |cos(7x)|)2
=
+ C.
14
=
= − cos(u) + C
= − cos(x2 ) + C.
Z
1
(−7 sin(7x)) dx
cos(7x)
dx =
2
2.
August 25, 2015
ex cos(ex )dx
Solution: Let u = ex . Then du = ex dx
and dx = du
. We have the following subex
stitution:
Z
Z
du
x
x
e cos(e )dx =
ex cos(u) x
e
Z
=
cos(u)du
4.
Z
x5
dx
6x2 − 22
Solution: Let u = 6x2 − 22. Then x2 =
u+22
and du = 12x dx. Thus
6
Z
x5
dx
6x2 − 22
Z
x(x2 )2 du
=
u 12x
Z u+22 2
1
6
=
du
12
u
Z 2
u + 44u + 222
1
=
du
432
u
Z
1
222
=
u + 44 +
du
432
u
2
1 u
2
=
+ 44u + 22 ln |u| + C.
432 2
= sin(u) + C
= sin(ex ) + C.
3. Assume 0 ≤ x ≤ π7 .
Z
tan(7x) ln |cos(7x)| dx
1
5. Find the area of the region bounded by
y = ln x2 , y = ln x, and x = e2 .You may
assume that
Z
ln(x) dx = x(ln(x) − 1) + C.
Thus we know that ln(x2 ) ≥ ln(x) for x in
the interval [1, e2 ] and the area between the
two curves is found by taking the difference
and integrating.
Solution: Let us begin by finding the
points of intersection. Set the first two equations equal to each other.
Z
e2
ln(x2 ) − ln(x) dx
AREA =
ln(x) = ln(x2 )
= 2 ln(x).
1
Z
e2
2 ln(x) − ln(x) dx
=
1
Z
Thus
=
0 = ln(x),
and
ln(x) dx
1
Z
and x = 1, y = 0 is the intersection pt.
Now we check the intersection with the third
equation.
ln((e2 )2 ) = 4
e2
e2
ln(x) dx
=
1
=
=
=
=
ln(e2 ) = 2
The first intersects the third at (e2 , 4), and
the second intersects the third at (e2 , 2)
2
2
x(ln(x) − 1)|e1
e2 (ln(e2 ) − 1) − 1(ln(1) − 1)
e2 (2 − 1) − 1(0 − 1)
e2 + 1
6. (Past Test question) Find the total 7. (Past Test question) Find the area of the
area enclosed by the graphs x = 8y 2 − y + y 3 region bounded by the graphs x = y + 1 and
and x = y 2 + 7y 3 .
x = y 2 − y − 7.
Solution: Let us equate the two equations
Solution: Let us equate the two equations
to find the intersection pts.
to identify their intersection pts.
2
3
2
3
8y − y + y = y + 7y
y + 1 = y2 − y − 7
0 = 6y 3 − 7y 2 + y
0 = y 2 − 2y − 8
= y(6y − 1)(y − 1)
0 = (y + 2)(y − 4)
Thus the two curves intersect when y =
0, 61 , 1. Evaluating each equation at the test
1
Thus our region is contained in the y, we find that
pt. 12
interval [−2, 4]. By evaluating each curve at
2 3
the pt y = 0, we see that the curve x = y +1
1
1
1
−
≈ −0.0272, and has larger x values on the interval [−2, 4]
8
+
12
12
12
than the curve y 2 − y − 7. Thus
2
3
Z 4
1
1
+7
≈ 0.0110.
y + 1 − (y 2 − y − 7) dy
Area =
12
12
−2
Z 4
Thus on the interval [0, 16 ], y 2 + 7y 3 is larger
=
−y 2 + 2y + 8 dy
than 8y 2 − y + y 3 . Evaluating each equation
−2
at the test pt. 12 , we find that
4
−y 3
y2
=
+
2
+
8y
2 3
3
2
−2
1
1
1
3
2
−
= 1.625, and
8
+
−4
4
2
2
2
=
+2 +8·4
3
2
2
3
(−2)2
−(−2)3
1
1
+2
+ 8 · (−2)
−
+7
= 1.125.
3
2
2
2
100
=
.
Thus on the interval [ 16 , 1], 8y 2 − y + y 3 is
3
larger than y 2 + 7y 3 . So the area between
the two curves is the sum of the two integrals integrals:
Z
1
6
Z 01
y 2 + 7y 3 − (8y 2 − y + y 3 ) dy
8y 2 − y + y 3 − (y 2 + 7y 3 ) dy.
1
6
3
√
2 − 0 dx +
2 − x − 1 dx
0
1
#5
"
3
2
2(x
−
1)
= 2x|10 + 2x −
3
Z
8. Find the area of the region in the
first√quadrant bounded by the graphs of
y = x − 1 and y = 2.
There are two ways to solve this problem.
You could either integrate over y or over x.
I will show both solutions.
1
Z
5
Area =
1
= (2 · 1 − 2 · 0)
3
2(5 − 1) 2
+ 2·5−
3
!
!
3
2(1 − 1) 2
− 2·1−
3
16
= 2 + 10 −
− (2 − 0)
3
14
=
.
3
Solution integrating over x: Let us find
the √
intersection of the lines y = 2 and
y = x−1
√
2 =
x−1
4 = x−1
5 = x.
Solution integrating over y: The interval of integration
is [0, 2]. The top function
√
is y = x − 1a and the bottom function is
x = 0. Solving for the top function as a
function of y, we find x = y 2 + 1. Thus
Z 2
y 2 + 1 dy
Area =
0
2
y3
=
+ y 3
0
23
+2
=
3
14
=
.
3
Since we are restricted to the first quadrant
our range of integration is [0, 5]. On the
interval from [0, 1] the bottom function is
y = 0 and √
on the interval [1, 5] the bottom
function is x − 1. y = 2 is the top function
on the entire interval. Thus
a
4
Remember to solve for x as a function of y