Notes: `y = cos(pix), y = 4x^2 - 1` Sketch the region enclosed by the given curves and find its area.

Tuesday, 15 September 2015

$\displaystyle{y}={\cos{{\left(\pi{x}\right)}}},{y}={4}{{x}}^{{2}}-{1}$ Sketch the region enclosed by the given curves and find its area.

$\displaystyle{y}={\cos{{\left(\pi{x}\right)}}},{y}={4}{{x}}^{{2}}-{1}$

Refer the attached image. Graph of cos(pix) is plotted in blue color and graph of y=4x^2-1 is plotted in red color.

From the graph , the curves intersect at x= $\displaystyle\pm$ 1/2.

Area enclosed by the curves A= $\displaystyle{\int_{{-\frac{{1}}{{2}}}}^{{\frac{{1}}{{2}}}}}{\left({\cos{{\left(\pi{x}\right)}}}-{\left({4}{{x}}^{{2}}-{1}\right)}\right)}{\left.{d}{x}\right.}$

$\displaystyle{A}={2}{\int_{{0}}^{{\frac{{1}}{{2}}}}}{\left({\cos{{\left(\pi{x}\right)}}}-{4}{{x}}^{{2}}+{1}\right)}{\left.{d}{x}\right.}$

$\displaystyle{A}={2}{{\left[\frac{{\sin{{\left(\pi{x}\right)}}}}{\pi}-{4}{\left(\frac{{{x}}^{{3}}}{{3}}\right)}+{x}\right]}_{{0}}^{{\frac{{1}}{{2}}}}}$

$\displaystyle{A}={2}{{\left[\frac{{\sin{{\left(\pi{x}\right)}}}}{\pi}-\frac{{{4}{{x}}^{{3}}}}{{3}}+{x}\right]}_{{0}}^{{\frac{{1}}{{2}}}}}$

$\displaystyle{A}={2}{\left({\left(\frac{{\sin{{\left(\frac{\pi}{{2}}\right)}}}}{\pi}-\frac{{4}}{{3}}{{\left(\frac{{1}}{{2}}\right)}}^{{3}}+\frac{{1}}{{2}}\right)}-{\left(\frac{{\sin{{\left({0}\right)}}}}{\pi}-\frac{{4}}{{3}}{{\left({0}\right)}}^{{3}}+{0}\right)}\right)}$

$\displaystyle{A}={2}{\left({\left(\frac{{1}}{\pi}-\frac{{4}}{{3}}{\left(\frac{{1}}{{8}}\right)}+\frac{{1}}{{2}}\right)}-{0}\right)}$

$\displaystyle{A}={2}{\left(\frac{{1}}{\pi}-\frac{{1}}{{6}}+\frac{{1}}{{2}}\right)}$

$\displaystyle{A}={2}{\left(\frac{{1}}{\pi}+\frac{{2}}{{6}}\right)}$

$\displaystyle{A}=\frac{{2}}{\pi}+\frac{{2}}{{3}}$