1
Spring 2017, Math 126
Quiz 2
Name: Iurii Posukhovskyi
Problem 1.
(2 points) Find the indenite integral
Z
cos(2x)ex dx.
Note that after integrating by parts twice with dv = ex dx we will get the same integral and, so, we
will be able to solve for it. Let u = cos(2x) and dv = ex dx, then du = −2 sin(2x)dx and v = ex . Integrating
by parts once we get
Z
Z
Solution.
cos(2x)ex dx = ex cos(x) + 2
sin(2x)ex dx.
Letting u = sin(2x) and dv = ex dx (du = 2 cos(2x)dx, v = ex ) , and integrating by parts one more time we
get
Z
x
Z
x
sin(2x)ex dx
Z
x
x
x
= e cos(2x) + 2 e sin(2x) − 2 cos(2x)e dx
Z
x
x
= e cos(2x) + 2e sin(2x) − 4 cos(2x)ex dx.
cos(2x)e dx = e cos(2x) + 2
So, we have
Z
x
x
Z
x
cos(2x)e dx = e cos(2x) + 2e sin(2x) − 4
Z
5
Z
Problem 2.
cos(2x)ex dx
cos(2x)ex dx = ex cos(2x) + 2ex sin(2x)
cos(2x)ex dx =
ex cos(2x) + 2ex sin(2x)
+ C.
5
(2 points) Find the indenite integral
Z
arctan(x)dx
Solution.
Integrating by parts with u = arctan(x) and dv = 1 · dx, so that du =
Z
Z
1 · arctan(x)dx = x arctan(x) −
x2
1
dx
x2 +1
and v = x, we get
x
dx.
+1
To evaluate the last integral we use the substitution u = x2 + 1, then du = 2xdx, and we get
Z
Z
x
dx
+1
Z
1
1
= x arctan(x) −
du
2
u
1
= x arctan(x) − ln |u| + C
2
1 = x arctan(x) − ln x2 + 1 + C.
2
1 · arctan(x)dx = x arctan(x) −
x2
2
Problem 3.
(2 points) Find the indenite integral
Z
Solution.
Here, the substitution u =
Z
√
cos( x)dx
√
x helps. With du =
√
cos( x)dx =
1
√
dx
2 x
we ge
√
2 x cos(u)du
Z
= 2 u cos(u)du
Z
Integrating by parts with v = u and dw = cos(u)du, so that dv = du and w = sin(u), we get
Z
√
cos( x)dx = 2
Z
u cos(u)du
Z
= 2 u sin(u) − sin(u)du
= 2u sin(u) + 2 cos(u) + C
√
√
√
= 2 x sin( x) + 2 cos( x) + C.
(4 points) Set up the integrals to compute the area enclosed between two curves. You don't
have to evaluate them.
Problem 4.
a) y = x3 and y = x.
b) y = (x − 2)2 and y = |x − 2|.
Solution.
a) Area =
Z
b) Area =
Z
0
3
Z
x − xdx +
−1
1
x − x3 dx.
0
2
2
Z
(2 − x) − (x − 2) dx +
1
2
3
(x − 2) − (x − 2)2 dx.