Exam I Study Guide
1.Antiderivative
If F 0 (x) = f (x), in MA223, we call f (x) the derivative of F (x); in MA224, we also call F (x)
an antiderivative of f (x). Just notice that all the antiderivatives of f (x) are in the form
F (x) + C, where C is an arbitrary constant, if F 0 (x) = f (x).
2.The Indefinite Integral
If F 0 (x) = f (x) as above, we write
Z
f (x)dx = F (x) + C,
and call both sides of the above equality the indefinite integral of f (x).
3.Rules for Integrating
Common Functions
Z
The constant rule:
kdx = kx + C for a constant k
Z
xn+1
+ C if n 6= −1
The power rule:
xn dx =
n+1
Z
1
dx = ln |x| + C for all x 6= 0
The logarithmic rule:
Z x
1
The exponential rule:
ekx dx = ekx + C for constant k 6= 0
k
4.Algebraic Rules for Indefinite
Integration
Z
Z
The constant multiple rule:
kf (x)dx = k f (x)dx for a constant k
Z
Z
Z
The sum rule: (f (x) + g(x))dx = f (x)dx + g(x)dx
Z
Z
Z
The difference rule: (f (x) − g(x))dx = f (x)dx − g(x)dx
Exercise Page 381: 12, 21
5.Initial Value Problems
If we are given f 0 (x) and one point (a, f (a)), we get f (x) by forming the indefinite integral
of f 0 (x) where C can be found using (a, f (a)).
Exercise Page 382: 33, 34, 38, 48; Page 395: 52
Z
6. Using substitution to integrate
f (x)dx
Most integrals are impossible to obtained directly. We can use substitution by letting u
to be some part of the integrand f (x). The aim of this method is to convert the original
1
complicated integral to a much simpler one so that we can use the formulae above. Here are
some guidelines:
(a) If possible, try to choose u so that u0 (x) is part of the integrand f (x).
(b) Consider choosing u as the part inside a radical, the denominator of a fraction, or the
exponent of an exponential function.
Z
1
dx, don’t choose u = x ln x, let u = ln x instead.
(c) Don’t ”oversubstitue”, e.g., in
x ln x
(d) If your choice of u doesn’t work. Don’t give up. Try another (and another) until you get
the right one.
Exercise Page 394-395: 3, 6, 12, 16, 17, 21, 25, 38, 31, 33
7.
the definite integral
Z b The Definite Integral Although closely related to each other,
Z
f (x)dx is fundamentally different than the indefinite integral f (x)dx. The former is a
a
real number while the latter is an expression having x as a variable and an arbitrary constant
C in the end. To evaluate an indefinite integral, we need the Fundamental Theorem of
Calculus:
b
Z b
f (x)dx = F (x)
a
a
0
where F (x) is an antiderivative of f (x), i.e., F (x) = f (x).
Z Usually we let F (x) be the part
having the constant C left out in the indefinite integral
b
that F (x) means F (b) − F (a).
f (x)dx = F (x) + C. Also notice
a
Exercise Page 410: 9, 10, 12
8. Evaluating the Definite Integral Using Substitution There are two methods for
doing definite integral using substitution. One changes the integral limits from a and b to
u(a) and u(b) along with the change of variables from x to u. The other evaluates the associated indefinite integral first and then use the fundamental theorem of calculus to get the real
number for the definite integral. Refer to the solution to Problem 1 of Quiz 2 posted online
for the detailed explanation. Note that DON’T confuse these two methods. Once you
picked your favorite, stick to it!
Exercise Page 411: 23, 24, 27
9. Algebraic Rules for Definite Integration
Definite integrals share the algebraic
rules of Zthe indefinite integrals:
Z
b
b
The constant multiple rule:
kf (x)dx = k
f (x)dx for a constant k
a
a
Z b
Z b
Z b
The sum rule:
(f (x) + g(x))dx =
f (x)dx +
g(x)dx
a
a
a
2
Z
b
Z
b
f (x)dx −
(f (x) − g(x))dx =
The difference rule:
Z
g(x)dx
a
a
a
b
Furthermore, definite integrals have another three rules which I never mentioned in class.
So
Z they won’t appear in Exam I. But it’s good to know them:
a
f (x)dx = 0 for all a.
Z a
b
f (x)dx.
f (x)dx = −
Za c
Z bb
Z c
f (x)dx =
f (x)dx +
f (x)dx: we can split an integral by splitting the interval
Za
a
a
b
whereas the sum rule above split the integral by splitting the integrand.
Exercise Page 411: 31, 38
10. Net Change It’s easy to see that another way to write the fundamental theorem of
calculus is
Z b
f (b) − f (a) =
f 0 (x)dx.
a
This is all about the net change problems in real life.
Exercise Page 412-413: 60, 64
11. Geometric
Z b Interpretation of the Definite Integral
f (x)dx mean geometrically? Well, if f (x) ≥ 0 on the interval a ≤ x ≤ b,
What does
a
Z b
f (x)dx is the area of the region on the x-y coordinate plane bounded above by the graph
a
of f (x), below by the x-axis (the graph of y = 0), left by the vertical line x = a and right
by the other vertical x = b. In general, given two functions f (x) and g(x), the area of
the region bounded by the graph of them is obtained by integrating the difference between
f (x) and g(x). If over the interval of intersection, f (x) ≥ g(x), we integrate f (x) − g(x);
if f (x) ≤ g(x), we integrate g(x) − f (x). But be careful! In some cases, the region
bounded have several components, among which the relationships between f (x)
and g(x) are different.
Exercise Page 411: 46; Page 427: 11, 12, 13
12. The Average Value of f (x)
The average value V of f (x) over a ≤ x ≤ b is simply
1
V =
b−a
Z
3
b
f (x)dx
a
Exercise Page 428-430: 24, 25, 49, 51
13. Consumer Willingness to Spend Given the demand function D(q), the consumer
total willingness to spend on q0 units of products is
Z q0
D(q)dq.
A(q0 ) =
0
14. Surpluses If q0 units are sold and D(q) is the consumers’ demand function for the
commodity, then the Consumers’ Surplus is
Z q0
D(q)dq − D(q0 )q0 .
CS =
0
If q0 units are sold and S(q) is the producers’ supply function for the commodity, then the
Producers’ Surplus is
Z q0
P S = S(q0 )q0 −
S(q)dq.
0
The equilibrium price pe is the common value for D(q) and S(q). So to get pe , we equate
D(q) = S(q) and then we get the solution for q. Plugging this q back in either one of D(q)
and S(q), we get pe .
Exercise Page 442: 1,2,8,11,15,16,18
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