Quiz 2 solution Math 1220

Quiz 2 solution
Math 1220-005 (Fall 2016)
Name(s):
Student ID Number(s):
1. Calculate
Sol.
Score:
R1
5
0 x2 −8x+20
R1
5
0 x2 −8x+20
dx. (2 pts)
dx = 5
R1
1
0 (x−4)2 +4
dx = 5
R −3
1
−4 u2 +4
du = 5 ·
1
2
−3
tan−1 ( u2 )
−4
= 52 (tan−1 (−3/2) − tan−1 (−2))
2. Calculate
Sol.
R
√2
3x 9x2 −1
R
√2
3x 9x2 −1
3. Calculate
R π/2
0
dx. (2 pts)
dx =
2
3
R
1
√ √
x· 9· x2 −1/9
dx =
2
9
·
1
1/3
|x|
sec−1 ( 1/3
) + C = 23 sec−1 (3|x|) + C.
cos(x) · sinh(sin x) dx. (2 pts)
Sol. Let u = sin(x), du = cos(x)dx.
R π/2
0
cos(x) · sinh(sin x) dx =
R1
0
1
sinh(u) du = cosh(u) =
0
cosh(1) − cosh(0) = cosh(1) − 1.
4. Find
R
t tan−1 (t) dt. (2 pts)
Sol. Integrate t but differentiate tan−1 (t).
=
t2
2
5. Find
Sol.
R
tan−1 (x) − 21 · (1 −
R1
0
R1
0
1
) dt
1+t2
=
t2
2
R
t tan−1 (t) dt =
t2
2
tan−1 (x) −
R
t2
2
·
1
1+t2
dt
tan−1 (x) − 2t + tan−1 (t) + C.
x2 ex dx. (2 pts)
1 R
R1
1
x2 ex dx = ex x2 − 0 ex · 2x dx = e − 2 · 0 ex · x. (**)
0
R1
x
To calculate for 0 e · x, we integrate by parts again, to get
1
e − ex = e − (e − 1) = 1.
R1
0
1 R
1
e · x = e · x − 0 ex dx =
x
x
0
Plug in this result to equation (**) above we get
R1
0
x2 ex dx = e − 2 · 1 = e − 2.
0