Let [t] denotes the greatest integer \( \le t \), then the value of \( \left[\int_{\pi}^{\frac{3\pi}{2}}\displaystyle x|\sin x|\,dx\right] \) is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 4
For \( \pi\lt x\lt\frac{3\pi}{2} \), sin x < 0, so \( |\sin x|=-\sin x \).
\( \displaystyle I=\int_{\pi}^{\frac{3\pi}{2}}x|\sin x|\,dx \)
\( \displaystyle =-\int_{\pi}^{\frac{3\pi}{2}}x\sin x\,dx \)
Integrating by parts, \( \int x\sin x\,dx=-x\cos x+\sin x \). So
\( \displaystyle I=\left[x\cos x-\sin x\right]_{\pi}^{\frac{3\pi}{2}} \)
\( \displaystyle =\left(\frac{3\pi}{2}\cdot 0-(-1)\right)-\left(\pi(-1)-0\right) \)
\( \displaystyle =1+\pi \)
Since \( \pi\approx 3.14 \), \( I\approx 4.14 \), and
\( \displaystyle [I]=[1+\pi]=4 \)