Notes: `y = sin(x), y = x, x = pi/2, x = pi` Sketch the region enclosed by the given curves. Decide whether to integrate with respect to `x` or `y`....

Sunday, 12 June 2016

$\displaystyle{y}={\sin{{\left({x}\right)}}},{y}={x},{x}=\frac{\pi}{{2}},{x}=\pi$ Sketch the region enclosed by the given curves. Decide whether to integrate with respect to $\displaystyle{x}$ or $\displaystyle{y}$ ....

$\displaystyle{y}={\sin{{\left({x}\right)}}},{y}={x},{x}=\frac{\pi}{{2}},{x}=\pi$

Refer the attached image. Graph of y=sin(x) is plotted in red color and y=x is plotted in blue color.

Area of the region enclosed by the given curves A= $\displaystyle{\int_{{\frac{\pi}{{2}}}}^{\pi}}{\left({x}-{\sin{{\left({x}\right)}}}\right)}{\left.{d}{x}\right.}$

$\displaystyle{A}={{\left[\frac{{{x}}^{{2}}}{{2}}-{\left(-{\cos{{\left({x}\right)}}}\right)}\right]}_{{\frac{\pi}{{2}}}}^{{\pi}}}$

$\displaystyle{A}={{\left[\frac{{{x}}^{{2}}}{{2}}+{\cos{{\left({x}\right)}}}\right]}_{{\frac{\pi}{{2}}}}^{{\pi}}}$

$\displaystyle{A}={\left(\frac{{{\left(\pi\right)}}^{{2}}}{{2}}+{\cos{{\left(\pi\right)}}}\right)}-{\left(\frac{{{\left(\frac{\pi}{{2}}\right)}}^{{2}}}{{2}}+{\cos{{\left(\frac{\pi}{{2}}\right)}}}\right)}$

$\displaystyle{A}={\left(\frac{{\pi}^{{2}}}{{2}}-{1}\right)}-{\left(\frac{{\pi}^{{2}}}{{8}}\right)}$

$\displaystyle{A}=\frac{{\pi}^{{2}}}{{2}}-\frac{{\pi}^{{2}}}{{8}}-{1}$

$\displaystyle{A}=\frac{{{3}{\pi}^{{2}}}}{{8}}-{1}\approx{2.7011}$