MATH 101 HOMEWORK 4 – SOLUTIONS
1. Find the area of the finite planar region bounded from above by the graph y = 3x − x2
and from below by y = |2x| − 6.
We first find the points of intersection. For x < 0, we have to solve 3x − x2 = −2x − 6,
ie. x2 − 5x − 6 = 0. This has 2 solutions, x = −1 and x = 6, the second one of which we
discard because 6 > 0. For x ≥ 0, we solve 3x − x2 = 2x − 6, ie. x2 − x − 6 = 0. Again,
this has two solutions, x = −2 and x = 3, but only the second one has x ≥ 0. So, there
are two intersection points, at x = −1 and x = 3.
The area in question is a sum of two areas, evaluated as follows:
Z
Z
0
0
2
(3x−x +2x+6)dx =
−1
Z
−1
Z
3
(3x − x − 2x + 6)dx =
3
0
and the total area is
0
19
x3 5x2
1 5
+
+6x = −( + +6) =
,
3
2
3 2
6
−1
(−x2 + x + 6)dx = −
2
0
(−x2 +5x+6)dx = −
3
27
x 3 x2
27 9
+
+ 6x = − + + 18 =
,
3
2
3
2
2
0
19 27
50
+
=
.
6
2
3
2. Evaluate the integrals:
Z 8
Z
512
1 1 512 −u
1 1
1 1
2
−x3
(a)
4−u − 512 .
x ·4
dx =
4 du =
=
3 1
3 − ln 4
3 ln 4 4 4
1
1
3
2
We substituted u = x , du = 3x dx.
(b) We first integrate by parts:
Z
−1
x(tan
x2
(tan−1 x)2 −
x) dx =
2
Z
2
x2 2 tan−1 x
x2
·
(tan−1 x)2 −
dx
=
2
1 + x2
2
Z
x2 tan−1 x
dx.
1 + x2
To evaluate the last integral, we substitute u = tan−1 x, x = tan u, du = (1 + x2 )−1 dx,
and get
x2
(tan−1 x)2 −
2
Z
x2
(tan−1 x)2 −
u tan u du =
2
2
x2
u2
=
(tan−1 x)2 +
−
2
2
Z
Z
(u sec2 u − u)du
x2 + 1
u sec u du =
(tan−1 x)2 −
2
2
Z
u sec2 u du.
The last integral is done by parts, and we get
x2 + 1
(tan−1 x)2 − u tan u +
2
Z
tan u du =
x2 + 1
(tan−1 x)2 − u tan u + ln | sec u| + C.
2
1
Now recall that u = tan−1 x, hence tan u = x, sec2 u = 1 + tan2 u. We now plug this in,
and get the final answer
p
x2 + 1
(tan−1 x)2 − x tan−1 x + ln 1 + x2 + C.
2
3. Evaluate the integrals (first make a substitution and then use integration by parts):
√
√
(a) We substitute u = x, du = dx/2 x:
Z
4 √
e
Z
x
4
dx =
1
√
√
2 xe
1
2
u
Z
= 2(ue −
1
x
dx
√ =
2 x
2
Z
2
2ueu du
1
2
u
e du) = 2(2e − e − e ) = 2e2 .
u
2
1
1
(b) Substitute x3 = u, 3x2 dx = du:
Z
1
x cos(x ) dx =
3
5
=
3
Z
u cos u du =
1
(u sin u − sin u)
3
1
1
(u sin u + cos u) + C = (x3 sin x3 + cos x3 ) + C.
3
3
2