Quiz
Math 121 Calculus II
D Joyce, Feb. 2013
Scale. 3 points for each problem, so 9 points in all. 8–9 A, 7–8 B, 5–6 C. Median 7.5.
Z
1
1. Evaluate the integral
(3x2 + 5x − 6) dx.
−1
An antiderivative of 3x2 + 5x − 6 is x3 + 52 x2 − 6x. By the ftc the integral is the
antiderivative evaluated between the limits of integration
1
x3 + 25 x2 − 6x = 13 + 25 · 12 − 6 · 1 − (−1)3 + 25 (−1)2 − 6(−1) = −10.
−1
Z
2. Use substitution to evaluate the indefinite integral
x2 cos(x3 + 1) dx
cos(x3 + 1) is a composition of functions, and that suggests u = x3 + 1 may be a useful
substitution. Then du = 3x2 dx, so the integral equals
Z
1
cos u du = 31 sin u + C = 13 sin(x3 + 1) + C
3
Note that an antiderivative of cos u is + sin u, not − sin u.
Be sure when you’re working with indefinite integrals that you always give your answer
in terms of the original variable.
Z
3. Use substitution to evaluate the definite integral
1
2e3x+1 dx
0
The substitution u = 3x + 1, du = 3 dx works here. Note that when x = 0, u = 1. Also,
when x = 1, u = 4. Therefore the integral is equal to
Z 4
4
2 u
e du = 23 eu 1 = 23 (e4 − e)
3
1
Note that an antiderivative of eu is itself, eu .
An alternate substitution that works for this integral is u = e3x+1 , du = 3e3x+1 dx. With
that substitution, the integral becomes
Z
e
e4
2
3
du =
2
3
e4
ue = 23 (e4 − e)
1