1549901369-Elements_of_Real_Analysis__Denlinger_

378 Chapter 7 • The Riemann Integral

Example 7.3.7 Use the technique of Theorem 7.3.6 to calculate

J 1
4
(x^2 - 4x + 5)dx.

Solution. Let f(x) = x^2 - 4x + 5. We know that f is integrable on [1,4]
since it is continuous there. Let Pn = { x 0 , x 1 , x2, · · · , Xn} be the partition that
subdivides [l, 4] into n subintervals of equal length. Then llPnll^4 ~^1 = ~'
and for i = 1 ,2,··· ,n,
3i 3
Xi = 1 + - and 6i = -.
n n

In each subinterval [xi-l, xi], choose the tag xi = Xi. Then
n
R(f, P~) = L f (xi)6i
i=l

n [ ( 3i )

2

=t; l+;- -4 ( l+;-3i ) +5; l^3

=~ n [ l+-+-6i 9i -4--+5 -

(^2 12) i ] 3
L.., n n^2 n n
i=l
-~^3 n [ 2--+-6i 9i
2
]
n L.., n n^2
i=l
3[n 5n gn]

• 2 2= 1 - - 2= i + 2 2= i^2
  n i=l n i=l n i=l
  = 2 [ 2 n _ ~ n(n + 1) + ~ n(n + 1)(2n + 1)]
  n n 2 n^2 6

6 _ 9. n +^1 + ~. n + 1. 2n +^1

n 2 n n

~ 6 - 9 + 9 6.

Now, llP~ll ~ 0, so by Theorem 7.3.6, R(f, P~) ~ f 1

4

f.

Therefore, .f 1

4

f = 6. D

*REGULAR PARTITIONS ARE SUFFICIENT

This material is here only for the curious. All others should

skip to Exercise Set 7.3.

*An asterisk with a theore m , proof, or other material in this chapter indicates that the

material is challenging and can b e omitted, especially in a one-semester course.