# f is not integrable on [ 0 , 1 ] .

### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

#### Solutions

Chapter 5.2, Problem 55E
To determine

## To show :   f is not integrable on [0,1] .

Expert Solution

### Explanation of Solution

Given information :

f(x)=0, xQ=1 xRQ

Proof :

f is bounded on [0,1] .

Let us take a partition P of [0,1]defined by P={x0,x1,........xn} where 0=x0<x1<x2<.....<xn=1

Let, M=x[0,1]supf(x) , m=xr[0,1]inff(xr)

Mr=x[xr1,xr]supf(x) , mr=x[xr1,xr]inff(x) for r=1,2,....n

Then M=1,m=o,Mr=1,mr=0 for r=1,2,....n

Now U(P,f)=M1(x1x0)+M2(x2x1)+.......+Mn(xnxn1)=1

L(P,f)=m1(x1x0)+m2(x2x1)+.......+mn(bxn1)=0

Let us consider the set [0,1] of all partitions of [0,1] .

The set L(P,f):P[0,1] of all partitions [0,1] .

The set, {L(P,f):P[0,1]} is the singleton set {0} .

So, upper bound of the set is 0 i.e. 01f=0

The set {U(P,f):P[0,1]} is the singleton set {1} .

The greatest lower bound of the set is 1 i.e. 01f=1

Since, _f01¯f01 , f is not integrable on [0,1] .

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