Here l=4

Half range sine series is given as

$f(x)=\Sigma_{n=1}^\infty b_n\sin \dfrac{n\pi x}{l}$

$\therefore f(x)=\Sigma_{n=1}^\infty b_n\sin \dfrac{n\pi x}{4}$

$Now,$

$b_n=\dfrac{2}{l}\int_0^lf(x).sin \dfrac{n\pi x}{l}dx$

$\therefore b_n=\dfrac{2}{4}\Big[\int_0^2x.sin \dfrac{n\pi x}{4}dx+\int_2^4(4-x).sin \dfrac{n\pi x}{4}dx\Big]$

$\therefore b_n=\dfrac{1}{2}\Bigg[\Big[(x)\Big(-\dfrac{4}{n\pi}.\cos \dfrac{n\pi x}{4}\Big)-(1)\Big(-\dfrac{16}{n^2\pi^2}.\sin \dfrac{n\pi x}{4}\Big)\Big]_0^2+ \Big[(4-x)\Big(-\dfrac{4}{n\pi}.\cos \dfrac{n\pi x}{4}\Big)-(-1)\Big(-\dfrac{16}{n^2\pi^2}.\sin \dfrac{n\pi x}{4}\Big)\Big]_2^4\Bigg]$

$\therefore b_n=\dfrac{1}{2}\Bigg[\Big[-\dfrac{8}{n\pi}.\cos \dfrac{n\pi}{2}+\dfrac{16}{n^2\pi^2}.\sin \dfrac{n\pi}{2}-0\Big]+\Big[0-\Big( -\dfrac{8}{n\pi}.\cos \dfrac{n\pi}{2}-\dfrac{16}{n^2\pi^2}.\sin \dfrac{n\pi}{2} \Big)\Big]\Bigg]$

$\therefore b_n=\dfrac{1}{2}\Big[\dfrac{32}{n^2\pi^2}.\sin \dfrac{n\pi}{2}\Big]$

$\therefore b_n=\dfrac{16}{n^2\pi^2}.\sin \dfrac{n\pi}{2}$

Therefore, Half range sine series is given as

$f(x)=\Sigma_{n=1}^\infty b_n\sin \dfrac{n\pi x}{l}$

$\therefore f(x)=\dfrac{16}{\pi^2}\Sigma_{n=1}^\infty \Big(\dfrac{1}{n^2}.\sin \dfrac{n\pi}{2}\Big)\sin \dfrac{n\pi x}{4}$