Calculus BC
Review Worksheet
8/4 8/6
1.
1
x 2 4
dx
x2 1
dx
x
2.
3.
16
x
4.
x
1
3
4 x 2 25 dx
1
2
dx
3 x 4
dx
x 6
5.
x
6.
3 x 2 4 x 4
x 3 2 x 2 dx
7.
2 x 2 x
( x 2)( x 2 1) dx
8.
sin x dx
1
9.
cos x dx
1
10.
11.
2
1
1 cos x
cos x 1
sin x 1 dx
sin x
s
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Lecture 4 : General Logarithms and Exponentials.
For a > 0 and x any real number, we dene
ax = ex ln a ,
a > 0.
The function ax is called the exponential function with base a.
Note that ln(ax ) = x ln a is true for all real numbers x and all a > 0. (We sa
Lecture 3 : The Natural Exponential Function: f (x) = exp(x) = ex
Last day, we saw that the function f (x) = ln x is one-to-one, with domain (0, ) and range (, ).
We can conclude that f (x) has an inverse function f 1 (x) = exp(x) which we call the natura
Lecture 11/12 : Partial Fractions
In this section we look at integrals of rational functions.
Essential Background
A Polynomial P (x) is a linear sum of powers of x, for example 3x3 + 3x2 + x + 1 or x5 x.
The degree of a polynomial P (x) is the highest po
Lecture 10 : Trigonometric Substitution
To solve integrals containing the following expressions;
a2 x 2
x 2 + a2
x 2 a2 ,
it is sometimes useful to make the following substitutions:
Expression
Substitution
2 x2
x = a sin , or = sin1 x
2
2
a
a
a2 + x 2
x =
Lecture 9 : Trigonometric Integrals
Mixed powers of sin and cos
Strategy for integrating
sinm x cosn xdx
We use substitution:
If n is odd use substitution with u = sin x, du = cos xdx and convert the remaining factors of cosine
using cos2 x = 1 sin2 x. Th
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Exam 2 Solutions.
1. Use Simpsons rule with n = 4 to estimate
ln 25 =
5
1
5
dx.
x
5
Solution: Here we have a = 1, b = 5, x = 51 = 1, and f (x) = x . Therefore, using the formula
4
for Simpsons rule, we have
5
5
15
5
5
5 5 1 5 20 10 20 5
dx
+4 +2 +4 +
=
Exam I Solutions
Math 10560, Spring 2014
1. (6 pts) The function
f (x) = x3 + x + ln(x)
is one-to-one (there is no need to check this). What is (f 1 ) (2)?
Solution: By guess and check we notice that f (1) = 2 so f 1 (2) = 1. Furthermore
1
x
f (x) = 3x2 +
Solutions to Exam 2, Math 10560
1. Which of the following expressions gives the partial fraction decomposition of the
function
3x2 + 2x + 1
f (x) =
?
(x 1)(x2 1)(x2 + 1)
Solution: Notice that (x2 1) is not an irreducible factor. If we write the denominato
MATH 10550 Assignment 4 1.4
Name:
1
1. A tank holds 1000 gallons of water, which drains from the bottom of the tank in half an
hour. The values in the table show the volume V of water remaining in the tank (in gallons)
after t minutes.
t (min)
5
10
15
20
Review Exam 2
Approximating an integral
Midpoint Rule If f is integrable on [a, b], then
n
b
f (x)dx Mn =
a
f (xi )x = x(f (x1 ) + f (x2 ) + + f (xn ),
i=1
Trapezoidal Rule If f is integrable on [a, b], then
b
f (x)dx Tn =
a
x
(f (x0 ) + 2f (x1 ) + 2f (x2
Name:
Date:
Worksheet 4,
Math 10560
Times indicate the amount of time that you would be expected to spend on the problem in
on an exam. All problems have appeared on old exams for Calculus 2.
2
x3 ln x dx.
1. (4 mins) Evaluate the integral
1
2. (4 mins) E
Name:
Instructor:
The following is the list of useful trigonometric formulas:
sin2 x + cos2 x = 1
1 + tan2 x = sec2 x
1
sin2 x = (1 cos 2x)
2
1
cos2 x = (1 + cos 2x)
2
sin 2x = 2 sin x cos x
1
sin(x y) + sin(x + y)
2
1
sin x sin y =
cos(x y) cos(x + y)
2
1.
1
x 2 4
I
2
= ln x x 4 - ln|2| + C
dx
1
2
x 4
1
2
= ln x x 4 + C
dx
dx
x 2 2 2
u=x
a=2
x = 2tan
dx = 2sec2d
1
I
2 2 d
sec
2
2
( 2 tan ) 2
2 2
sec
4 tan 2 4
2 2
sec
2. I
I
d
x2 1
dx
x
x 2 12
dx
x
u x
a 1
u a
sec
x 1
sec
dx 1 tan d
sec
sec 2 12
ta
Math 10260 Exam 2 Solutions Fall 2012.
1. This series is not initially a geometric series, but if we write
2n + (1)n 2 n 1 n
=
+
,
3n
3
3
n=0
n=0
n=0
then the two terms on the right hand side are both geometric series with the form a
rn with
n=0
a
|r| <