Question

Find the area of the figure bounded by the following curves
y = sin $x_2,y=2x/\pi ,$

Answer 1

$y=sinx$, $y=\frac{2x}{\pi }$

$sinx=\frac{2x}{\pi }$

$\frac{\pi }{2}sinx=x$

$sinx$ varies from $0$ to $1$

So, $x=0$ or $x=\frac{\pi }{2}$

$\frac{\pi }{2}$ Area under curve $={\int }_{x_1}^{x_2}(y_2-y_1)dx$

$={\int }_0^{\frac{\pi }{2}}(sinx-\frac{2x}{\pi })dx$

$=-cosx-\frac{x^2}{\pi }{]}_0^{\frac{\pi }{2}}$

$=-cos\frac{\pi }{2}-\frac{{\pi }^2}{4\pi }+cos0+\frac{0}{\pi }$

$=1-\frac{\pi }{4}$ Sq units