Page 1 | © 2014 by Janice L. Epstein 6.1: The Definite Integral CHAPTER 6: INTEGRATION
6.1: The Definite Integral
Summation (Sigma) Notation
If am, am+1, …, an are real numbers and m and n are integers such that
m £ n , then
n
åa = a
i
m
+am+1 +  + an-1 + an
where i is the index of summation.
i=m
6
Example: Evaluate
1
å
i =1 i + 1
8
Example: Expand
åx
k +1
k =5
1 1 1
1
1
Example: Write 1- + - + in sigma notation
4 9 16 25 36
25
Example: Evaluate
å7
i=3
Page 2 | © 2014 by Janice L. Epstein 6.1: The Definite Integral THEOREM: If c is a constant, then
n
(a)
å1 = n
(d)
i =1
n
(b)
n
å ca =cå a
i
i=m
n
(c)
i
i=m
n
n
å(a  b ) =å a  å b
i
n (n + 1)
2
i =1
n
n (n + 1)(2n + 1)
2
i
=
(e) å
6
i =1
2
n
é
ù
n
n
1
+
(
)
ú
(f) å i 3 = ê
ê
ú
i =1
ë 2 û
n
i
i=m
i
i=m
i
i=m
Example: Evaluate the following sums
100
(i)
å 6i
i =1
n
(ii)
å i (i +1)
i =1
æ1
1 ö÷
ç
(iii) å ç ÷÷
ç
è
ø
1
i
i
+
i =1
99
3
ù
1 éêæç i ÷ö
(iv) lim å êç ÷÷ + 1úú
ç
n¥
i =1 n êëè n ø
úû
n
åi =
Page 3 | © 2014 by Janice L. Epstein 6.1: The Definite Integral Area
Example: What is the area under the curve y = -2 x + 2 from 0 to 1?
Approximate with rectangles:
2
2
1
1
1
2
1
1
1
Page 4 | © 2014 by Janice L. Epstein 6.1: The Definite Integral More generally, partition the interval [ a, b ] as P = {a = x0 , x1 ,, xn = b} .
A subinterval [ xi-1 , xi ] has a width Dxi = xi - xi-1 and the point xi* is in
that subinterval.
The length of the longest subinterval is the norm of P,
P = max {Dx0 , Dx1 ,, Dxn }
The area of the ith rectangle is
Ai = f ( xi* )Dxi .
The total approximate area is the sum
n
S P = å f ( xi* )Dxi
i =1
The value of the sum SP will depend on the partition and is called a
Riemann sum of f on [ a, b ]
A Riemann sum is an approximation of the signed area between a function
f ( x) and the x-axis on a given interval. So areas above the x-axis are
positive and areas below the x-axis are negative.
Typically find either the left sum (by using the value of the function at the
left endpoint), the right sum (by using the value of the function at the
right endpoint), or the midpoint sum (by using the value of the function at
the midpoint of the sub-interval).
n
We say the area of the region is A = lim å f ( xi* )Dxi
P 0
i =1
Page 5 | © 2014 by Janice L. Epstein 6.1: The Definite Integral Example: You are given a function f, an interval, partition points, and a
description of the point xi* within the ith subinterval.
(a) Find P
(b) Sketch the graph of f and the approximating rectangles.
(c) Find the sum of the approximating rectangles.
(i) f ( x) = 16 - x 2 , [0, 4], P = {0, 1, 2, 3, 4}, xi* = left endpoints.
é pù
ì p p p pü
(ii) f ( x) = 4cos x, ê 0, ú , P = ïí0, , , , ïý , xi* = right endpoints
êë 2 úû
îïï 6 4 3 2 þïï
(iii) f ( x) = ln x and interval [1, 3]. Create the partition by using 3
subintervals of equal length and xi* = midpoints.
Page 6 | © 2014 by Janice L. Epstein 6.1: The Definite Integral Example: For the following functions, set up the area as a limit of a sum.
Do not evaluate the limit.
(i) f ( x) = x 2 + 3 x - 2 on the interval [1, 4] using right endpoints.
(ii) f ( x) = x 2 + 1 on the interval [0, 5] using left endpoints.
Example: Approximate the area under f ( x) = x 2 on the interval [1,3]
using a right Riemann sum with 4 rectangles of equal width. Then find
the area using n rectangles and let n   .
Page 7 | © 2014 by Janice L. Epstein 6.1: The Definite Integral Definite Integral
If f is a function defined on a closed interval [a, b], let P be a partition of
[a, b] with partition points x0, x1, … xn where a = x0 < x1 < … < x n = b
Choose points xi* in the subinterval [xi-1,xi] and let ∆xi = xi – xi-1 and ||P|| =
max {∆xi }. Then the definite integral of f from a to b is
b
ò
n
f ( x) dx = lim å f ( xi* )Dxi
P 0
a
i =1
if this limit exists. If the limit exists for f, then f is Riemann integrable
on the interval [a, b]. All continuous functions are Riemann integrable.
If f is positive on [a, b], then then the definite integral is the area under f
and above the x-axis from x=a to x=b. If f is not always positive, the
definite integral is the net area.
b
Midpoint Rule:
ò
a
where Dx =
n
f ( x) dx » å f ( xi )Dx =Dx éê f ( x1 ) +  + f ( xn )ùú
ë
û
i =1
x - xi
b-a
and xi = i-1
2
n
5
Example: Use the midpoint rule with n=4 to approximate
ò
1
x 2 + 1 dx
Page 8 | © 2014 by Janice L. Epstein 6.1: The Definite Integral Example: Use geometric formulas to evaluate the following definite
integrals
3
(i)
ò (1- 2 x) dx
0
3
(ii)
ò
x - 2 dx
-1
0
(iii)
ò
-2
4 - x 2 dx
Page 9 | © 2014 by Janice L. Epstein 6.1: The Definite Integral Properties of Definite Integrals
b
i.
ò c dx = c (b - a)
a
b
ii.
b
ò cf ( x) dx = c ò
a
b
iii.
a
b
ò ( f ( x)  g ( x)) dx = ò
a
a
iv.
ò
vi.
a
a
f ( x) dx = -ò f ( x) dx
a
b
c
b
b
f ( x) dx = ò f ( x) dx + ò f ( x) dx
a
vii.
f ( x) dx  ò g ( x) dx
f ( x) dx = 0
ò
ò
b
a
a
b
v.
f ( x) dx
a
c
If m £ f ( x) £ M for all x in the interval [ a, b ], then
b
m (b - a ) £ ò f ( x) dx £ M (b - a )
a
1
Example: Find
ò
x 5 + x 2 + 1 dx
1
Example: Given
3
ò
1
f ( x) dx = 4 and
3
3
1
1
ò g ( x) dx = -3, find ò éë f ( x) + 2 g ( x)ùû dx
Page 10 | © 2014 by Janice L. Epstein 5
Example: Write
ò
-3
0
6.1: The Definite Integral 6
f ( x) dx - ò f ( x) dx + ò f ( x) dx as a single integral
-3
5
2
Example: Find an upper and lower bound on
ò
x3 + 1 dx
0
Example: Express the following limits as definite integrals
5
n
1 n
1
2 çæ çæ 2i ÷ö
÷÷ö
(ii) lim å çç3ç1 + ÷÷ - 6÷
(i) lim å
2
n¥ n
n¥
÷ø
nø
i =1 1 + i
i =1 n ç
è çè
n
( )