11sir yeh kaiise kiya[7]
9 Answers
11same answer to nahin aa raha
but{x}=x-1 (1,2)
{x}=x-2 (2,3)
{x}=x-3 (3,4)
[x]=1 (1,2)
[x]=2 (2,3)
[x]=3 (3,4)
62this is simpler than it seems it will be
\int_{0}^{1}{x^{0}}dx+\int_{0}^{1}{x^{1}}dx+\int_{0}^{1}{x^{2}}dx+\int_{0}^{1}{x^{3}}dx
62think thoda sa ...
whe x is in between 1 to 2
{x} is in between 0 to 1
and [x] is 1
same logic goes for other intervals...
also dx = d{x}
62yup i made one small extra interval.. the first one..
there will be 3 summations...
\int_{0}^{1}{x^{1}}dx+\int_{0}^{1}{x^{2}}dx+\int_{0}^{1}{x^{3}}dx
Ans: 1/2+1/3+1/4
24felt the same.......but didnt have the courage to say that ur answer is wrong........[1][1]
11answer theek to aa raha hai
sir ke tarike se
SIR TUSSI TAAN GR8 HO
A VERY PRODUCTIVE METHOD.........
24haan theek to hai.............ab tujhe kya problem hai???????????[7][12][11]