1guees that's absolutely correct.
{x} ka differentiation kaise karen koi bataye ga kya?????
\int_{0}^{\pi /4}{\left\{2sinxcosx \right\}}d\left(x-\left[x \right] \right) where [x] and{x} r G.I.F and fractional part respectively.
ans-----> 1/2
-
UP 0 DOWN 0 1 2
2 Answers
29\left\{2sinxcosx \right\} = \left\{sin2x \right\}....x - \left[x \right] = \left\{x \right\}
since in the interval ( 0 , π/4 ) ..{sin2x} varies b/w 0 to 1 ...so the integral can be written as
\int_{0}^{\pi /4}{\left\{sin2x \right\}d\left\{x \right\}} = \int_{0}^{1}{tdt} = \frac{t^{2}}{2} = \frac{1}{2}
please correct me if i am wrong
1