EE 503, UNIVERSITY OF SOUTHERN CALIFORNIA, 2016
1
Review of Basic Calculus
Michael J. Neely
University of Southern California
http://www-rcf.usc.edu/∼mjneely
Abstract
These notes review basic calculus, including the chain rule, product rule, and the fundamental theorems of
calculus. The rules for integration by substitution and integration by parts are derived as consequences. Finally,
worked examples that apply the substitution rule and the integration by parts rule are given. The functions f (x)
considered in these notes are assumed to be real valued functions defined over x ∈ R, that is, f : R → R.
I. BASIC CALCULUS RESULTS
A. Differentiation
Let f (x), g(x) be differentiable functions and let f 0 (x), g 0 (x) be the derivatives.
0
0
0
• Chain rule: If h(x) = f (g(x)) then h (x) = f (g(x))g (x).
•
Product rule: If h(x) = f (x)g(x) then h0 (x) = f 0 (x)g(x) + f (x)g 0 (x).
B. Fundamental theorems of calculus
Rb 0
•
f (x)dx = f (b) − f (a). [Assuming f 0 (x) is a continuous function]
a
•
d
dx
R x
a
f (t)dt = f (x). [Assuming f (t) is a continuous function]
II. I NTEGRATION TECHNIQUES
A. Substitution rule for integration
If f (u), g(x), g 0 (x) are continuous functions and a, b are real numbers, then
Z g(b)
Z b
0
f (g(x))g (x)dx =
f (u)du
g(a)
a
Proof: (Application of chain rule for differentiation) Define h(x) = F (g(x)), where F (x) is an
antiderivative of f (x). By the chain rule we get:
h0 (x) = f (g(x))g 0 (x)
Integrating both sides gives:
Z
b
0
Z
h (x)dx =
a
b
f (g(x))g 0 (x)dx
a
On the other hand, we have:
Z b
Z
(a)
(b)
(c)
0
h (x) = h(b) − h(a) = F (g(b)) − F (g(a)) =
a
g(b)
f (u)du
g(a)
where (a) holds by the fundamental theorem of calculus applied to the continuous function h0 (x), (b)
holds by definition of h, and (c) holds by the fundamental theorem of calculus applied to the continuous
function f (u).
EE 503, UNIVERSITY OF SOUTHERN CALIFORNIA, 2016
2
B. Integration by parts
If f (x), f 0 (x), g(x), g 0 (x) are continuous functions and a, b are real numbers, then
Z b
Z b
b
0
f 0 (x)g(x)dx
f (x)g (x)dx = f (x)g(x) | a −
a
a
f (x)g(x) | ba
where
= f (b)g(b) − f (a)g(a).
Proof: (Application of product rule for differentiation) Define h(x) = f (x)g(x). The product rule
gives:
h0 (x) = f 0 (x)g(x) + f (x)g 0 (x)
Integrating both sides gives:
Z
b
Z
0
h (x)dx =
Z
Z
0
f (x)g(x)dx +
a
On the other hand,
b
a
b
b
f (x)g 0 (x)dx
a
(a)
(b)
h0 (x)dx = h(b) − h(a) = f (b)g(b) − f (a)g(a)
a
where (a) holds by the fundamental theorem of calculus applied to the continuous function h0 (x), and (b)
holds by definition of h.
C. Examples of integration via the substitution rule
Consider the equation for the substitution rule:
Z g(b)
Z b
0
f (u)du
f (g(x)) g (x)dx =
|{z} | {z }
g(a)
a
u
du
The substitution rule can be implemented by performing the change of variables
u = g(x)
du = g 0 (x)dx
and then remembering to appropriately change the integration limits.
Rb
2
1) Solving a xex dx: Define u = x2 , du = 2xdx. Then:
Z b
Z b2
1 u
1
1 2
2
2
x2
xe dx =
e du = eu |ba2 = (eb − ea )
2
2
a
a2 2
R3 x
x
x
x
2) Solving 1 e cos(e )dx: Define u = e , du = e dx. Then:
Z 3
Z e3
x
x
e cos(e )dx =
cos(u)du = sin(e3 ) − sin(e)
1
e
D. Examples of integration by parts
Consider the integration by parts equation:
Z b
Z b
b
0
g(x) f 0 (x)dx
f (x) g (x)dx = f (x) g(x) | a −
|
{z
}
|{z}
|{z}
|{z}
a |{z} | {z }
a
u
dv
u
v
v
Integration by parts can be implemented by the change of variables:
u = f (x) , dv = g 0 (x)dx
du = f 0 (x)dx , v = g(x)
du
EE 503, UNIVERSITY OF SOUTHERN CALIFORNIA, 2016
1) Solving
R5
1
3
x cos(x)dx: Define u = x, dv = cos(x)dx. Then:
u = x , dv = cos(x)dx
du = dx , v = sin(x)
So:
Z
5
x sin(x)|51
x cos(x)dx =
1
Z
−
5
sin(x)dx
1
= x sin(x)|51 + cos(x)|51
= 5 sin(5) + cos(5) − sin(1) − cos(1)
2) Solving
R∞
0
xe−x dx: Define u = x, dv = e−x dx. Then:
u = x , dv = e−x dx
du = dx , v = −e−x
So:
Z
∞
−x
xe dx =
0
−xe−x |∞
0
Z
−
∞
−x
Z
(−e )dx = 0 +
0
0
∞
e−x dx = −e−x | ∞
0 = 1