Math 247 Winter 2002
Solutions
Quiz 5
1. Use separation of variables to solve the following dierential equations:
dy
a)
= cos(2x), y (0) = 3
dx
Solution: We separate variables:
dy
= cos(2x)
dx
=
dy =
Math 247 Winter 2002
Solutions
Quiz 4 (Worksheet)
b
ex dx for b = 1, b = 10 and b = 1000.
1. Compute
0
b
b
ex dx = ex + C , so
Solution: Well,
ex dx = ex
0
= eb e0 =
0
1 eb .
For b = 1 this is 1 e1 0.
Math 247 Winter 2002
Solutions
Quiz 3
1. Use integration by parts to evaluate the indenite integrals:
a)
x cos(x) dx
Solution: Let u = x and v = cos(x), so u = 1 and v = sin(x). Integration by parts s
Math 247 Winter 2002
Solutions
Quiz 2
1. Compute the average value of f (x) = 4 x2 on the interval [2, 2].
1
Solution: The average value of a function f on the interval [a, b] is
ba
Here f (x) = 4 x2
Math 247
Solutions to Review for Final
10
sin(t2 + 1) dt. Compute F (x).
1. Let F (x) =
x2
h(x)
f (t) dt, then F (x) = f (h(x)h (x)
Solution: By Leibnizs Rule, if F (x) =
g (x)
f (g (x)g (x). We have
Math 247
Review for Final
10
sin(t2 + 1) dt. Compute F (x).
1. Let F (x) =
x2
2. Compute the average value of f (x) = ln(x) over the interval [1, e2 ].
3. Compute the area between the curves y = x2 +
Math 247
Review for Exam #2
Solutions
1. Use integration by parts to evaluate the integrals:
a)
x sec2 (x) dx
Solution: Let u = x and v = sec2 (x). Then u = 1 and v = tan(x). We get
x sec2 (x) dx = x
Math 247
Solutions to Review for Exam #1
4
1. Write the sum in expanded form:
k (k + 1)
k=1
4
Solution: Well,
k (k + 1) = 1(1 + 1) + 2(2 + 1) + 3(3 + 1) + 4(4 + 1).
k=1
2
4 x2 dx
2. Use geometry to co