c
Math 151, Benjamin
Aurispa
6.1 Sigma Notation
The Greek letter,
P
, is used to represent a sum of many terms:
If am , am+1 , . . . , an are real numbers and m and n are integers with m < n, then
n
X
i=m
ai = am + am+1 + am+2 + · · · + an
This form is called sigma notation and the letter i is called the index of summation.
Example: Calculate
5
X
(i2 + i).
i=2
Example: Calculate
200
X
200
X
2
i=1
Example: Write the sum
Example: Evaluate
i=4
2
5
+
3
6
+
25 X
1
i=5
2
4
7
1
−
i−1
i
+
5
8
+ ... +
18
21
i=1
using sigma notation.
Properties of Sums Using Sigma Notation:
(a)
n
X
cai = c
n
X
(ai ± bi ) =
i=m
(b)
n
X
ai
i=m
i=m
n
X
i=m
ai ±
n
X
n
X
bi
i=m
1
a
c
Math 151, Benjamin
Aurispa
Special Sums (Do not memorize):
(a)
n
X
i=
i=1
(b)
n
X
i2 =
n(n + 1)(2n + 1)
6
i3 =
i=1
(c)
n
X
n(n + 1)
2
i=1
Calculate
n(n + 1)
2
10
X
2
i
i=1
Calculate
10
X
i2
i=3
Calculate
n
X
i=1
(i − 4)(i + 1)
Calculate lim
n→∞
n
X
2
i=1
"
2i
+
n n
2i
n
3 #
2
c
Math 151, Benjamin
Aurispa
6.2 Area
Goal: Suppose we have a function f (x) where f (x) ≥ 0 on the interval [a, b]. We want to be able to find
the area under the curve between x = a and x = b.
We can estimate the area by dividing up the region into intervals and then forming rectangles. The area
can then be approximated by the sum of the areas of these rectangles. The more rectangles, the better the
approximation.
3
c
Math 151, Benjamin
Aurispa
Method:
1. Determine Width of the Rectangles
Create a partition P of the interval [a, b] by dividing the interval into n smaller subintervals.
The x-values we choose to divide the interval into subintervals are called the partition numbers and denoted
x0 , x1 , x2 , . . . , xn .
a
b
. . .
x0
x1
x2
x3
[x0 , x1 ], [x1 , x2 ], [x2 , x3 ], [x3 , x4 ]
...
x4
x n−2
x n−1
xn
[xn−2 , xn−1 ], [xn−1 , xn ]
Notes:
The first partition number should always be a and the last partition number should always be b. However,
the subintervals do not have to be equally spaced.
The lengths of these subintervals will be the widths of our rectangles.
The length of the ith subinterval is denoted ∆xi , where ∆xi = xi − xi−1
The norm of the partition, ||P ||, is defined to be the largest ∆xi . ||P || = max{∆x1 , ∆x2 , . . . , ∆xn }.
4
6
9
14
17
2. Determine Height of the Rectangles
Choose a number within each subinterval [xi−1 , xi ]. We will call this number x∗i . This number can be the
left endpoint, right endpoint, midpoint, or any other point in the subinterval.
We choose the function value at this point, f (x∗i ), to be the height of the rectangle over that interval.
3. Determine Area of the Rectangles
The area of the rectangle corresponding to the subinterval [xi−1 , xi ] is now f (x∗i )∆xi .
So, the total area of all the rectangles is
n
X
f (x∗i )∆xi . This is called a Riemann Sum.
i=1
4
c
Math 151, Benjamin
Aurispa
Example: Consider the function f (x) = 20 − x2 on the interval [0, 4]. Approximate the area under the
curve on this interval by using the partition P = {0, 2, 3, 4} and choosing x∗i to be the left endpoint of each
subinterval.
If we want n EQUALLY-SPACED subintervals for an interval [a, b], what is ∆xi ?
Example: Consider the function f (x) = x2 + 1 on the interval [2, 10]. Approximate the area under the curve
on this interval by using 4 equal-length subintervals and choosing x∗i to be the midpoint of each subinterval.
5
c
Math 151, Benjamin
Aurispa
Example: Consider the function f (x) = x3 on the interval [0, 3]. Approximate the area under the curve on
this interval by using 6 equal-length subintervals and choosing x∗i to be the right endpoint of each subinterval.
EXACT Area: These Riemann sums are just an approximation for the area under the curve. As the the
widths of these rectangles get smaller and smaller (and we thus have more and more rectangles), we will get
closer and closer to the actual area.
A = lim
||P ||→0
n
X
f (x∗i )∆xi
i=1
If the intervals all have the same length, this limit can be expressed as
A = lim
n→∞
n
X
f (x∗i )∆xi
i=1
In general then, the exact area under the graph of a curve f (x) on an interval [a, b] can be found by computing
the limit:
n
n
A = lim
n→∞
X
i=1
f (x∗i )∆xi = lim
n→∞
X
f (a + i∆xi ) ∆xi
i=1
Note: We could have also used the left endpoint or midpoint instead of the right endpoint to develop the
above formula, but it ultimately doesn’t make a difference since we are taking the limit.
6
c
Math 151, Benjamin
Aurispa
Set up the limit to find the exact area under the graph of f (x) =
√
Set up the limit to find the exact area under the graph of f (x) =
1
+ sin(x) on the interval [3, 7].
x2
7
x3 + 1 on the interval [1, 4].
c
Math 151, Benjamin
Aurispa
6.3 The Definite Integral
We saw in the previous section that if f (x) ≥ 0 on an interval [a, b], then the exact area under the graph
between x = a and x = b is
n
A = lim
||P ||→0
X
f (x∗i )∆xi
i=1
We now define the definite integral of f from a to b as the above limit:
b
Z
f (x) dx = lim
||P ||→0
a
If the subintervals are equally spaced, then ∆xi =
Z
f (x∗i )∆xi
i=1
b−a
and this limit can be represented as:
n
b
a
n
X
f (x) dx = lim
n→∞
n
X
f (x∗i )∆xi
i=1
If the limit exists, then f is said to be integrable on this interval.
Notes: In the notation ab f (x) dx, f (x) is called the integrand and a and b are call the limits of integration. (a is the lower limit and b is the upper limit).
R
Once again, if f (x) ≥ 0 on the interval, then the definite integral can be interpreted as the area under the
graph of f between x = a and x = b
*If f is not always positive, the definite integral is still defined, but now represents the NET area.*
For the graph of f below, compute the following definite integrals using the indicated areas.
Z
A
f (x) dx =
0
20
Z
B
Z
C
Z
D
Z
C
10
f (x) dx =
A
40
B
15
C
D
A
f (x) dx =
0
f (x) dx =
A
f (x) dx =
C
How much total area is bounded between the curve and the x-axis between x = 0 and x = D?
8
c
Math 151, Benjamin
Aurispa
Express the following limit as a definite integral on the interval [0, 2].
lim
n→∞
" n
X
2i 6
2
i=1
n
n
2i
−4
n
#
The Midpoint Rule for definite integrals means to approximate the integral by using a midpoint Riemann
Sum (just as in 6.2).
Use the Midpoint Rule with n = 4 to approximate
Z
4
−4
(x2 − 4) dx.
Evaluate the following definite integrals by interpreting each in terms of area.
(1)
Z
6
−1
(2x − 4) dx
9
c
Math 151, Benjamin
Aurispa
(2)
(3)
Z
Z
0
5p
25 − x2 dx
5
−2
|x − 3| dx
Properties of the Integral:
(1)
(2)
(3)
(4)
(5)
(6)
Z
Z
Z
Z
Z
Z
b
c dx = c(b − a) where c is any constant
a
b
a
[f (x) ± g(x)] dx =
b
cf (x) dx = c
a
c
f (x) dx +
Z
b
a
f (x) dx ±
b
f (x) dx = −
a
Z
b
g(x) dx
a
b
f (x) dx
a
b
f (x) dx =
Z
b
f (x) dx
a
c
a
a
Z
Z
Z
a
f (x) dx
b
f (x) dx = 0
a
Example: If
Z
2
5
f (x) dx = 7 and
Z
2
5
g(x) dx = −4, calculate
10
Z
5
2
(3f (x) + g(x)) dx.
c
Math 151, Benjamin
Aurispa
Example: Write the following as a single integral.
Z
3
−2
Z
f (x) dx −
0
f (x) dx +
−2
Z
5
f (x) dx
3
More Properties:
(1) If f (x) ≥ 0 for all x on [a, b], then
Z
b
f (x) dx ≥ 0
a
(2) If f (x) ≥ g(x) for all x on [a, b], then
Z
b
f (x) dx ≥
a
Z
b
g(x) dx
a
(3) If m ≤ f (x) ≤ M for all x on [a, b], then
m(b − a) ≤
Z
b
a
f (x) dx ≤ M (b − a)
Suppose a continuous function f (x) has an absolute maximum value of 12 and an absolute minimum value
of 3 on the interval [2, 8]. What can be said about the value of
Z
8
2
6.4 The Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus, Part 2:
If f is continuous on [a, b], then
Z
b
a
f (x) dx = F (b) − F (a)
where F is any antiderivative of f .
Evaluate the following definite integrals.
Z
1
3
3
1
−
3
x
x
dx
11
f (x) dx ?
c
Math 151, Benjamin
Aurispa
Z
2
Z
4
Z
π
4
0
(3ex − 2x ) dx
1
u2 − 1
√ du
u u
(sec2 θ + 2 cos θ) dθ
0
Find the area under the curve f (t) = (t2 − t)2 between t = −2 and t = 0.
12
c
Math 151, Benjamin
Aurispa

 √ 1
Evaluate
f (x) dx where f (x) =
1 − x2

−1/2
Z
π/2
sin x
− 12 ≤ x < 0
0≤x≤
π
2
Find the area under the curve f (x) = |x2 − 4| between x = 0 and x = 4.
If f (8) = 12,
f′
is continuous, and
Z
8
−1
f ′ (x) dx = 9, what is f (−1)?
13
c
Math 151, Benjamin
Aurispa
Application: Suppose an object has velocity function v(t). Then, on the time interval from t = a to t = b:
Z
b
a
v(t) dt = s(b) − s(a)
by the 2nd Part of the Fundamental Theorem since the position function s(t) is an antiderivative of v(t).
This is the displacement of the object on the time interval.
The actual total distance traveled by the object is
Z
b
a
|v(t)| dt. Why?
Example: Suppose the velocity of an object is given by the function v(t) = t2 − 2t − 3. Find the displacement
and total distance traveled by the object on the time interval 0 ≤ t ≤ 4.
The indefinite integral is used to indicate the process of finding the most general antiderivative of f (x):
Z
f (x) dx = F (x) + C
where F (x) is an antiderivative of f (x) (i.e. F ′ (x) = f (x)).
Find the general indefinite integral
Z √
5
+ csc x cot x − 3 x dx
2
1+x
14
c
Math 151, Benjamin
Aurispa
The Fundamental Theorem of Calculus, Part 1:
If f is continuous on [a, b], then the function g(x) defined by
g(x) =
Z
x
f (t) dt
a
a≤x≤b
is continuous on [a, b] and differentiable on (a, b) and g′ (x) = f (x).
In other words,
Let g(x) =
decreasing?
Z
d
dx
Z
x
f (t) dt = f (x)
a
x
f (t) dt, where f is graphed below. Calculate g(0), g(2), and g(6). Where is g(x) increasing?
0
g represents the “area so far”. The FTC Part 1 says that g is an antiderivative of f since g ′ (x) = f (x).
Calculate the derivatives of the following functions.
Z
x
Z
x3
Z
x
Z
12
Z
x2
g(x) =
cos x
g(x) =
Z
g(x) =
t2 dt
1
g(x) =
t2 dt
sin x
g(x) =
−7
g(x) =
x
p
3
6 − t2 dt
p
3
6 − t2 dt
ln u
du
+3
u2
3
√
x
tan t
dt
t2
15