Given, f:[0,1]→R is integral. So, f:[0,1]→R is also Riemann integral.

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
Section: Chapter Questions
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For this question, why f: [0,1]→R is also Riemann integral if f:[0,1]→R is integral? What theorem is being used?
A set AC [0, 1] is dense in [0, 1] iff every open interval that intersects [0, 1] contains a point of A. Suppose
f : (0, 1] → R is integrable and f(x) = 0 for all x € A with A dense in [0, 1]. Show that
| f(x) dx = 0.
Transcribed Image Text:A set AC [0, 1] is dense in [0, 1] iff every open interval that intersects [0, 1] contains a point of A. Suppose f : (0, 1] → R is integrable and f(x) = 0 for all x € A with A dense in [0, 1]. Show that | f(x) dx = 0.
Step1
a)
Given, f:[0,1]→R_ is integral.
So, f:[0,1]→R is also Riemann integral.
Then all its Riemann sum converge to
Then consider one particular Riemann sum that is,
For the subdivision {0=x, <x <x <....
X, = 1},
%3D
In each subinterval [x,,x], we take one point ɛ, EA.
Step2
b)
Then the Riemann sum is always zero that is,
n-1
i-0
So, its limit 0 to 1.
Therefor,
Hence proved.
Transcribed Image Text:Step1 a) Given, f:[0,1]→R_ is integral. So, f:[0,1]→R is also Riemann integral. Then all its Riemann sum converge to Then consider one particular Riemann sum that is, For the subdivision {0=x, <x <x <.... X, = 1}, %3D In each subinterval [x,,x], we take one point ɛ, EA. Step2 b) Then the Riemann sum is always zero that is, n-1 i-0 So, its limit 0 to 1. Therefor, Hence proved.
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