Answer
The region bounded by
$y= 2x ,\ \ y=\sin x ,\ \ x=0,\ \ x=\pi/2 $
rotates about $x=-1$
Work Step by Step
Given
$$
\int_0^{\pi / 2} 2 \pi(x+1)(2 x-\sin x) d x
$$
Compare with
$$V= \int_a^b 2\pi r(x) h(x)dx$$
Here $r(x)= x+1,\ \ h(x)=2x-\sin x$. This represents the volume of the generated solid when the region bounded by
$$y= 2x ,\ \ y=\sin(x) ,\ \ x=0,\ \ x=\pi/2 $$
rotates about $x=-1$