# Chapter 6: Integration ```Chapter 6: Integration
Section 6.1: The Definite Integral
Definition: If m and n are integers such that m ≤ n, then
n
X
ak = am + am+1 + am+2 + &middot; &middot; &middot; + an−1 + an .
k=m
The term k is called the index of summation, m is called the lower limit of the sum,
and n is called the upper limit of the sum.
Example: Write each sum in expanded form.
(a)
6
X
k=1
(b)
8
X
1
k+1
xk
k=5
Example: Write each sum in sigma notation.
√
√
√
√
√
(a) 3 + 4 + 5 + 6 + 7
(b) 1 −
1 1
1
1
1
+ −
+
−
4 9 16 25 36
1
Theorem: (Summation Formulas)
Suppose that c is a constant and n is a positive integer. Then
1.
n
X
cak = c
k=m
2.
n
X
n
X
(ak &plusmn; bk ) =
k=m
3.
n
X
ak
k=m
n
X
ai &plusmn;
k=m
n
X
bk
k=m
c = cn
k=1
4.
n
X
k=
k=1
5.
n
X
n(n + 1)
2
k2 =
k=1
n(n + 1)(2n + 1)
6
Example: Evaluate each sum.
(a)
6
X
k(k + 2)
k=3
(b)
4
X
2k+1
k=1
2
(c)
100
X
4k
k=1
(d)
n
X
(5 − 2k)
k=1
(e)
n
X
(2k + 3)2
k=1
3
Area Under a Curve
Suppose that f is a continuous function on the interval [a, b]. What is the area of the
region S that lies under the curve y = f (x) from x = a to x = b?
To approximate the area of this region, we begin by subdividing the interval [a, b] into n
subintervals by choosing partition points x0 , x1 , x2 , . . . , xn such that
a = x0 &lt; x1 &lt; x2 &lt; &middot; &middot; &middot; &lt; xn−1 &lt; xn = b.
Then the n subintervals are
[x0 , x1 ], [x1 , x2 ], . . . , [xn−1 , xn ].
This subdivision is called a partition of [a, b], and denoted by P . The length of the kth
subinterval [xk−1 , xk ] is
∆xk = xk − xk−1 .
The length of the longest subinterval is called the norm of P and denoted by ||P ||. That is,
||P || = max{∆x1 , ∆x2 , . . . , ∆xn }.
Then we choose a representative point x∗k in each subinterval [xk−1 , xk ] and construct an
approximating rectangle Rk with base ∆xk and height f (x∗k ).
The area of the region S is approximated by the Riemann sum
A≈
n
X
f (x∗k )∆xk .
k=1
4
Example: Estimate the area under the graph of f (x) = 16 − x2 on [0, 4] using four equal
subintervals and left endpoints. Is your estimate an underestimate or overestimate?
5
Example: Estimate the area under the graph of f (x) = 4 cos x on [0, π/2] using four equal
subintervals and right endpoints. Is your estimate an underestimate or overestimate?
6
The Definite Integral
To obtain a better approximation of the area of the region under the curve y = f (x) on
the interval [a, b], we choose finer and finer partitions of the interval [a, b] and evaluate the
Riemann sum.
Definition: If f is a continuous function defined on a closed interval [a, b], let P be a partition
of [a, b] with partition points x0 , x1 , x2 , . . . , xn , where
a = x0 &lt; x1 &lt; x2 &lt; &middot; &middot; &middot; &lt; xn = b.
Choose representative points x∗k in [xk−1 , xk ] and let ∆xk = xk −xk−1 and ||P || = max{∆xk }.
Then the definite integral of f from a to b is
Z
b
f (x) dx = lim
||P ||→0
a
n
X
f (x∗k )∆xk ,
k=1
provided that this limit exists. If the limit does exist, then f is called integrable on [a, b].
The function f (x) is called the integrand and a and b are called the limits of integration.
Note: To evaluate a definite integral as the limit of a Riemann sum, it is often easier to use a
regular partition. That is, a partition for which all the subintervals have the same length,
b−a
.
n
∆x =
If we choose x∗k to be the right endpoint, then
x∗k
= a + k∆x = a +
b−a
n
k.
Then the definite integral of f from a to b is
Z
b
f (x) dx = lim
a
n→∞
7
n
X
k=1
f (x∗k )∆x.
Z
Example: Evaluate
4
(x2 + 3x − 2) dx as the limit of a Riemann sum.
1
8
Note: If f (x) ≥ 0 on [a, b], then the definite integral of f from a to b represents the area
under the graph of y = f (x) on [a, b]. However, if f (x) ≤ 0 on, then the definite integral of
f from a to b is the negative of the area above y = f (x) on [a, b].
Example: Evaluate each definite integral by interpreting it in terms of signed areas.
Z 2√
(a)
4 − x2 dx
−2
Z
3
(2 − x) dx
(b)
−1
Z
3
|3x − 6| dx
(c)
0
9
Theorem: (Properties of the Definite Integral)
Suppose that f and g are integrable functions on [a, b] and c is a constant.
Z b
c dx = c(b − a)
1.
a
Z
b
Z
b
f (x) dx
cf (x) dx = c
2.
a
a
Z
b
b
Z
[f (x) &plusmn; g(x)] dx =
3.
a
Z
b
Z
f (x) dx &plusmn;
g(x) dx
a
a
a
f (x) dx = 0
4.
a
Z
b
a
Z
f (x) dx = −
5.
f (x)dx
a
Z
6.
b
b
c
Z
f (x) dx =
b
Z
f (x) dx +
a
a
f (x) dx
c
7. If f (x) ≤ g(x) for a ≤ x ≤ b, then
Z
b
Z
f (x) dx ≤
a
b
g(x) dx.
a
8. If m ≤ f (x) ≤ M for a ≤ x ≤ b, then
b
Z
m(b − a) ≤
f (x) dx ≤ M (b − a).
a
Z
Example: Evaluate
1
x2 cos x dx.
1
10
Example: Write each expression as a single definite integral.
Z 7
Z 10
f (x) dx
f (x) dx −
(a)
2
2
Z
5
Z
0
f (x) dx −
(b)
−3
Z
f (x) dx +
−3
6
f (x) dx
5
Example: Find an upper and lower bound for the definite integral
Z
2
√
x3 + 1 dx.
0
11
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