I Exam Study Guide
1.Antiderivative
If F (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 (x) = f (x).
2.The Indenite Integral
If F (x) = f (x) as above, we write
f (x)dx = F (x) + C,
and call both sides of the above equality the indenite integral of f (x).
3.Rules for Integrating Common Functions
The constant rule:
k dx = kx + C for a constant k
xn dx =
The power rule:
The logarithmic rule:
The exponential rule:
xn+1
+ C if n = 1
n+1
1
dx = ln |x| + C for all x = 0
x
1
ekx dx = ekx + C for constant k = 0
k
4.Algebraic Rules for Indenite Integration
The constant multiple rule:
The sum rule:
k f (x)dx = k
(f (x) + g (x))dx =
The dierence rule:
f (x)dx for a constant k
f (x)dx +
(f (x) g (x))dx =
g (x)dx
f (x)dx
g (x)dx
Exercise Page 381: 12, 21
5.Initial Value Problems
If we are given f (x) and one point (a, f (a)), we get f (x) by forming the indenite integral
of f (x) where C can be found using (a, f (a)).
Exercise Page 382: 33, 34, 38, 48; Page 395: 52
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 u (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.
1
dx, dont choose u = x ln x, let u = ln x instead.
(c) Dont oversubstitue, e.g., in
x ln x
(d) If your choice of u doesnt work. Dont 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 Denite Integral Although closely related to each other, the denite integral
b
f (x)dx is fundamentally dierent than the indenite 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 indenite integral, we need the Fundamental Theorem of
Calculus:
b
b
f (x)dx = F (x)
a
a
where F (x) is an antiderivative of f (x), i.e., F (x) = f (x). Usually we let F (x) be the part
having the constant C left out in the indenite integral
f (x)dx = F (x) + C . Also notice
b
means F (b) F (a).
that F (x)
a
Exercise Page 410: 9, 10, 12
8. Evaluating the Denite Integral Using Substitution There are two methods for
doing denite 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 indenite integral rst and then use the fundamental of theorem calculus to get the real
number for the denite integral. Refer to the solution to Problem 1 of Quiz 2 posted online
for the detailed explanation. Note that DONT confuse these two methods. Once you
picked your favorite, stick to it!
Exercise Page 411: 23, 24, 27
9. Algebraic Rules for Denite Integration
Denite integrals share the algebraic rules of the indenite integrals:
b
The constant multiple rule:
b
The sum rule:
b
kf (x)dx = k
a
b
(f (x) + g (x))dx =
a
f (x)dx for a constant k
a
b
f (x)dx +
a
g (x)dx
a
2
b
b
b
f (x)dx
(f (x) g (x))dx =
The dierence rule:
g (x)dx
a
a
a
Furthermore, denite integrals have another three rules which I never mentioned in class.
So they wont appear in Exam I. But its good to know them:
a
f (x)dx = 0 for all a.
a
a
b
f (x)dx =
a
f (x)dx.
b
b
c
f (x)dx =
c
f (x)dx +
a
a
f (x)dx: we can split an integral by splitting the interval
b
whereas the sum rule above split the integral by splitting the integrand.
Exercise Page 411: 31, 38
10. Net Change Its easy to see that another way to write the fundamental theorem of
calculus is
b
f (b) f (a) =
f (x)dx.
a
This is all about the net change problems in real life.
Exercise Page 412-413: 60, 64
11. Geometric Interpretation of the Denite Integral
b
f (x)dx mean geometrically? Well, if f (x) 0 on the interval a x b,
What does
b
a
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 dierence 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 dierent.
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=
ba
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
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
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
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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