Math 247, Midterm I.
0. Write your name here:
3
1. Find
x-1 x dx.
1/3
x-1 x -1 dx = (x-2/3 - x-1 )dx = We have x dx = x 3 C. 3 x - ln |x| + 3 x-1 3 Answer: x dx = 3 x - ln |x| + C.
3
x1/3 1/3
- ln |x| + C =
2. Approximate We have Answer:
Math 247
Homework #3
(5.3) #5. Find the area of the region bounded by the line and curve
y = x2 + 1, y = 4x 2 (in the rst quadrant).
Solution: First we have to nd where the graphs cross. To do this we solve x2 + 1 =
4x 2 = x2 4x + 3 = 0 = (x 3)(x 1) = 0.
Math 247
Homework #4
(6.2) #1. Use integration by parts to evaluate
x cos(x) dx
Solution: Let
u =x
u =1
v = cos(x)
v = sin(x)
We get
x cos(x) dx =
= x sin(x)
1 sin(x) dx
= x sin(x)
sin(x) dx
= x sin(x) cos(x) + C
= x sin(x) + cos(x) + C
(6.2) #2. Use in
Math 247
Homework #5
(6.4) #3. Explain why the integral is improper and evaluate it.
2
dx
1 + x2
0
Solution: The integral is improper since its over an unbounded interval. We evaluate:
0
2
dx = lim
b
1 + x2
b
0
2
dx = lim 2 arctan(x)
b
1 + x2
= lim 2 arct
Math 247
Solutions
Midterm 1
3
1. Write the sum in expanded form:
k=1
3
1
1
1
1
=
+
+
k (k + 1)
1(1 + 1) 2(2 + 1) 3(3 + 1)
Solution: Well,
k=1
3
2. Given that
1
k (k + 1)
5
5
f (t) dt = A,
1
f (t) dt = B and
f (t) dt = C , compute the following:
1
2
2
a)
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 , a = 2 and b = 2, so the average value is
1
2 (2)
=
1
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 says,
x cos(x) dx
= x sin(x)
1 sin(x) dx
= x sin(x)
si
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 0.632.
For b = 10 this is 1 e10 0.999954.
For b = 1000 th
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 =
=
dy = cos(2x) dx
=
cos(2x) dx
y=
1
sin(2x) + C
2
Now w
Math 247
Review for Exam #1
4
1. Write the sum in expanded form:
k (k + 1)
k=1
2
4 x2 dx
2. Use geometry to compute the denite integral:
2
0
5
5
3. Given that
f (t) dt = 3,
0
f (t) dt = 1, compute the following:
f (t) dt = 2 and
0
1
1
a)
f (t) dt
0
2
b)
f
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 compute the denite integral:
0
Solution: The integral rep
Math 247
Review for Exam #2
1. Use integration by parts to evaluate the integrals:
a)
x sec2 (x) dx
b)
xe3x dx
c)
x2 sin(x) dx
d)
arctan(x) dx
e)
x arctan(x) dx
f)
x cos(x) dx
0
e
x5 ln(x) dx
g)
1
2
x3 ex dx
h)
2. Determine if the following integrals conv
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 tan(x)
b)
1 tan(x) dx = x tan(x) ln | sec(x)| + C
xe3x
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 + 1 and y = x + 3.
4. Evaluate the integrals: (you may wa
Math 247
Homework #2
(5.2) #7. Find
dy
where y =
dx
x
tan(u) du, 0 < x <
x
.
4
Solution: By Leibnizs Rule (page 255)
x
d
tan(u) du = tan(x) (x) tan(x) (x) = tan(x) (1) tan(x) (1) =
dx x
tan(x) + tan(x) = tan(x) tan(x) = 0
dy
(5.2) #8. Find
where y =
dx
x3
University of Oregon: Math 247
Final Exam, June 9, 2011
No Calculators, Notes, or Formula Sheets. Show all your work or you may get no credit
Do all your work on the paper provided. You have 2 hours
1. State carefully the Fundamental Theorem of Calculus,
University of Oregon: Math 246
Quiz 1, April. 18, 2011
No Calculators, Notes, or Formula Sheets. Show all your work or you may get no credit!
Do all your work on the paper provided. You have 15 minutes
1. State the Fundamental Theorem of Calculus, part II
University of Oregon: Math 247
Quiz 2, May 9, 2011
No Calculators, Notes, or Formula Sheets. Show all your work or you may get no credit!
Do all your work on the paper provided. You have 20 minutes
1. Finish this theorem: If f is continuous on [a, b] and
University of Oregon: Math 247
Quiz 3, June 1, 2011
No Calculators, Notes, or Formula Sheets. Show all your work or you may get no credit!
Do all your work on the paper provided. You have 15 minutes
1. Solve the following dierential equation:
dy
1
= (3 y
University of Oregon: Math 247
Test 1, April 21, 2010
No Calculators, Notes, or Formula Sheets. Show all your work or you may get no credit
Do all your work on the paper provided. You have 50 minutes
1. State the Fundamental Theorem of Calculus, Part II.
University of Oregon: Math 247
Test 2, May 5, 2011
No Calculators, Notes, or Formula Sheets. Show all your work or you may get no credit
Do all your work on the paper provided. You have 50 minutes
1. State carefully:
(a) The Fundamental Theorem of Calculu
University of Oregon: Math 247
Test 3, May 19, 2011
No Calculators, Notes, or Formula Sheets. Show all your work or you may get no credit
Do all your work on the paper provided. You have 50 minutes
1. State carefully:
(a) The Fundamental Theorem of Calcul
Math 247
Homework #6
(7.1) #1. Solve the pure time dierential equation.
dy
= x + sin(x) where y0 = 0 for x0 = 0
dx
Solution: We separate variables:
dy
= x + sin(x)
=
dy = x + sin(x) dx
dx
=
dy =
=
x + sin(x) dx
y=
x2
cos(x) + C
2
Now we apply the initial
Math 247
Solutions
Midterm 1
3
1. Write the sum in expanded form:
k=1
3
1
1
1
1
=
+
+
k (k + 1)
1(1 + 1) 2(2 + 1) 3(3 + 1)
Solution: Well,
k=1
3
2. Given that
1
k (k + 1)
5
5
f (t) dt = A,
1
f (t) dt = B and
f (t) dt = C , compute the following:
1
2
2
a)
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 , a = 2 and b = 2, so the average value is
1
2 (2)
=
1
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 says,
x cos(x) dx
= x sin(x)
1 sin(x) dx
= x sin(x)
si
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 0.632.
For b = 10 this is 1 e10 0.999954.
For b = 1000 th