MATH 126
QUIZ 8
Wed. Dec. 7, 2016
NAME (please print legibly):
1. (14 pts) Evaluate the following indefinite integrals.
Z
1
(a) (3x2 − 4x−9 + 6x−1 )dx = x3 + x−8 + 6 ln |x| + C
2
Z
(b)
sec2 t dt = tan(t) + C
√
4
Z
(c)
Z
x1/4 dx = 6 · 45 x5/4 + C
6 x dx = 6
Z
Z
(2x + 1)(x − 3)dx =
(d)
Z
u+4
du =
u
Z
e−23x dx =
(e)
(f)
Z
(2x2 − 5x − 3)dx = 23 x3 − 25 x2 − 3x + C
(1 + 4u−1 )du = u + 4 ln |u| + C
1
e−23x
−23
+C
Z
(g)
cos(t/12)dt = 12 sin(t/12) + C
2. (8
definite integrals.
Z 4pts) Evaluate the following
4
3
(a)
(x2 + 2x)dx = 13 x3 + x2 = ( 43 + 16) − ( 13 + 1)
1
1
Z
ln 3
(b)
(c)
1
Z
x
0
0
Z
ln 3
e dx = e = eln 3 − e0 = 3 − 1 = 2
x
5
5
1
dx = ln |x| = ln 5 − ln 1 = ln 5
x
1
π/2
(d)
sin xdx = 0 (since sin x is odd).
−π/2
3. (10
Z pts) Evaluate the following integrals using the substitution method.
(a)
2x(1 + x2 )100 dx. Let u = 1 + x2 , du = 2xdx. We get:
Z
u100 du =
(1 + x2 )101
u101
+C =
+ C.
101
101
Z
(b)
x
dx. Make the same substitution as in the previous problem. Then x dx = 12 du,
1 + x2
so we get:
Z
1
1
du = 12 ln |u| + C = 12 ln |1 + x2 | + C.
2
u
Z
(c)
esin(x) cos(x)dx. Let u = sin x, du = cos x dx. Then we get
Z
Z
(d)
x4 sin(x5 )dx. Let u = x5 , du = 5x4 dx. Then we get:
1
5
Z
(e)
eu du = eu + C = esin x + C.
Z
1
1
sin u du = − cos u + C = − cos(x5 ) + C.
5
5
π/2
cos3 (x) sin(x)dx. Let u = cos x, du = − sin x dx. When x = 0, u = cos(0) = 1.
0
When x = π/2, u = cos(π/2) = 0. The integral becomes
Z 0
Z 1
u4 1 1
3
−
u du =
u3 du = = .
4 0 4
1
0
4. (0 pts) (FOR PRACTICE ONLY)
Take the DERIVATIVE of the following functions.
(a) f (x) = 23 x3 + πx2 − e3
f 0 (x) = 2x2 + 2πx.
(b) g(x) = ln(2x3 + 3x + 1)
g 0 (x) =
(c) h(x) =
2x + 1
x2 − 1
h0 (x) =
6x2 + 3
.
2x3 + 3x + 1
(x2 − 1)(2) − (2x + 1)(2x)
.
(x2 − 1)2
(d) A(x) = e3x sin x
A0 (x) = e3x cos x + 3e3x sin x.
(e) B(x) = cos(sin(tan(x)))
B 0 (x) = − sin(sin(tan(x))) cos(tan(x)) sec2 (x).