question evaluate the integral below. ∫₀^(π/2) x sin(2x) dx be sure to place the argument of any…

Explanation:

Step1: Apply integration - by - parts formula

The integration - by - parts formula is $\int_{a}^{b}u\mathrm{d}v=uv|{a}^{b}-\int{a}^{b}v\mathrm{d}u$. Let $u = x$ and $\mathrm{d}v=\sin(2x)\mathrm{d}x$. Then $\mathrm{d}u=\mathrm{d}x$ and $v=-\frac{1}{2}\cos(2x)$. [ \begin{align*} \int_{0}^{\frac{\pi}{2}}x\sin(2x)\mathrm{d}x&=\left[- \frac{1}{2}x\cos(2x)\right]{0}^{\frac{\pi}{2}}-\int{0}^{\frac{\pi}{2}}\left(-\frac{1}{2}\cos(2x)\right)\mathrm{d}x\ \end{align*} ]

Step2: Evaluate $\left[- \frac{1}{2}x\cos(2x)\right]_{0}^{\frac{\pi}{2}}$

[ \begin{align*} \left[- \frac{1}{2}x\cos(2x)\right]_{0}^{\frac{\pi}{2}}&=-\frac{1}{2}\times\frac{\pi}{2}\cos(\pi)-\left(-\frac{1}{2}\times0\times\cos(0)\right)\ &=-\frac{\pi}{4}\times(- 1)-0\ &=\frac{\pi}{4} \end{align*} ]

Step3: Evaluate $\int_{0}^{\frac{\pi}{2}}\left(-\frac{1}{2}\cos(2x)\right)\mathrm{d}x$

Let $t = 2x$, then $\mathrm{d}t = 2\mathrm{d}x$. When $x = 0$, $t = 0$; when $x=\frac{\pi}{2}$, $t=\pi$. [ \begin{align*} \int_{0}^{\frac{\pi}{2}}\left(-\frac{1}{2}\cos(2x)\right)\mathrm{d}x&=-\frac{1}{4}\int_{0}^{\pi}\cos(t)\mathrm{d}t\ &=-\frac{1}{4}[\sin(t)]_{0}^{\pi}\ &=-\frac{1}{4}(\sin(\pi)-\sin(0))\ &=0 \end{align*} ]

Step4: Combine the results

[ \begin{align*} \int_{0}^{\frac{\pi}{2}}x\sin(2x)\mathrm{d}x&=\frac{\pi}{4}-0\ &=\frac{\pi}{4} \end{align*} ]

Answer:

$\frac{\pi}{4}$