integral definite doubt

\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

2 Answers

\left\{2sinxcosx \right\} = \left\{sin2x \right\}....x - \left[x \right] = \left\{x \right\}

since in the interval ( 0 , Ï€/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

guees that's absolutely correct.

{x} ka differentiation kaise karen koi bataye ga kya?????

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