CSIR NET MATHEMATICS | Real Analysis | Riemann Integration DEC 2011

Name: CSIR NET MATHEMATICS | Real Analysis | Riemann Integration DEC 2011
Uploaded: 2026-10-14T04:34:29.080Z
Duration: 4 min 2 s
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Oct 14, 2026

CSIR NET MATHEMATICS | Real Analysis | Riemann Integration DEC 2011
Real Analysis, Riemann Integration,

Example (marks 3)
Define f :[0,1]  [0,1] by f(x)=(2^k-1)/2^k for x∈[(2^(k-1)-1)/2^(k-1) ,(2^k-1)/2^k ],k≥1.
Then f is a Riemann-integral function such that

CSIR NET Answer Key December 2011,
CSIR NET Answer Key June 2019, Mathematical Science,
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Welcome friends, in this video we are going to discuss the problem of real analysis from CSIR NET December 2011. Example define f from closed interval 0,1 to closed interval 0,1 by fx is equal to 2 power k minus 1 divided by 2 power k. For x belongs to closed interval 2 power k minus 1 minus 1 divided by 2 power k minus 1, 2 power k minus 1, 2 power k, for k natural then equal to 1. Then f is a Riemann integrable function such that we have four options. First option, integration 0 to 1 fx dx equal to 2 by 3. Second option, 1 by 2 less than integration 0 to 1 fx dx less than 2 by 3. Third option, limit 0 to 1 fx dx is equal to 1. And fourth option, 2 divided by 3 less than integration 0 to 1 fx dx less than 1. Now let's say problem of Riemann integral. So first we recall Riemann integration. integration 0 to 1 fx dx is equal to limit n tends to infinity sigma k equal to 1 to n f of x k into delta x for the our function f of x k is equal to 2 power k minus 1 divided by 2 power k and delta x is the difference of this partition interval 2 power k minus 1 divided by 2 power k minus 2 power k minus 1 minus 1 divided by 2 power k minus 1. Now in the last term 2 power k minus 1 can be written in the form of 2 power k divided by on the denominator 2 power k divided by 2. Now if we simplify then we have 2 power k minus 2 divided by 2 power k. Now if we subtract this two number then we have 1 upon 2 power k. So we can write down limit n tends to infinity sigma k equal to 1 to n 2 power k minus 1 divided by 2 power 2k if we separate the denominator values then 1 upon 2 power k minus 1 upon 2 power 2k so this summation will become a summation up to the infinity sigma k equal to 1 to infinity 1 upon 2 power k minus sigma k equal to 1 to infinity 1 upon 4 power k because 2 power 2k is equal to 4 power k Both are the geometric series so we recall the sum of geometric series which is sigma k equal to 1 to infinity a into r power k minus 1 has value a upon 1 minus r where a is a first term of the series and r is a ratio. In our example first summation is a geometric series with the ratio 1 by 2 so r equal to 1 by 2 and the first term is 1 by 2 therefore a upon 1 minus r is equal to 1 by 2 divided by 1 minus 1 by 2 and in the second summation ratio is 1 by 4 and the first term is also 1 by 4 so 1 upon 4 divided by 1 minus 1 upon 4 then we have 1 minus 1 by 3 and which is equal to 2 by 3 which is the option number one so option number one is correct so thanks for watching for more videos like share subscribe press the bell icon of Next Gen Maths