int_{0}^{pi /2} { x - [sin x] } ,dx is equal | vedclass

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(D) We are given the integral $I = \int_{0}^{\pi /2} \{ x - [\sin x] \} \,dx$. Since $\{ x \} = x - [x]$,we can rewrite the expression as $\{ x - [\si

Vedclass · Accepted Answer

None of these

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(D) We are given the integral $I = \int_{0}^{\pi /2} \{ x - [\sin x] \} \,dx$. Since $\{ x \} = x - [x]$,we can rewrite the expression as $\{ x - [\sin x] \} = x - [\sin x] - [x - [\sin x]]$. However,for $x \in [0, \pi/2]$,we know that $0 \le \sin x Thus,the integral simplifies to $I = \int_{0}^{\pi /2} \{ x - 0 \} \,dx = \int_{0}^{\pi /2} \{ x \} \,dx$. Since $0 \le x \le \pi/2$ and $\pi/2 \approx 1.57$,the interval $[0, \pi/2]$ contains the integer $1$. We split the integral: $I = \int_{0}^{1} x \,dx + \int_{1}^{\pi /2} (x - 1) \,dx$. Calculating the first part: $\int_{0}^{1} x \,dx = [\frac{x^2}{2}]_{0}^{1} = \frac{1}{2}$. Calculating the second part: $\int_{1}^{\pi /2} (x - 1) \,dx = [\frac{x^2}{2} - x]_{1}^{\pi /2} = (\frac{\pi^2}{8} - \frac{\pi}{2}) - (\frac{1}{2} - 1) = \frac{\pi^2}{8} - \frac{\pi}{2} + \frac{1}{2}$. Adding them together: $I = \frac{1}{2} + \frac{\pi^2}{8} - \frac{\pi}{2} + \frac{1}{2} = \frac{\pi^2}{8} - \frac{\pi}{2} + 1$. Since this result is not among the options,the correct choice is $D$.

$x$	$1$	$2$	$3$	$4$
$y$	$0.7111$	$0.7222$	$0.7333$	$0.7444$

$\int_{0}^{\pi /2} \{ x - [\sin x] \} \,dx$ is equal to

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