Blog: `y = (1/4)x^2, y = 2x^2, x + y = 3, x>0` Sketch the region enclosed by the given curves and find its area.

Friday, September 4, 2015

$\displaystyle{y}={\left(\frac{{1}}{{4}}\right)}{{x}}^{{2}},{y}={2}{{x}}^{{2}},{x}+{y}={3},{x}>{0}$ Sketch the region enclosed by the given curves and find its area.

Given $\displaystyle{y}=\frac{{1}}{{4}}{{x}}^{{2}},{y}={2}{{x}}^{{2}},{x}+{y}={3},{x}>{0}$

Find the intersection point of $\displaystyle{y}={2}{{x}}^{{2}}$ and $\displaystyle{y}=-{x}+{3}.$

$\displaystyle{2}{{x}}^{{2}}=-{x}+{3}$

$\displaystyle{2}{{x}}^{{2}}+{x}-{3}={0}$

$\displaystyle{\left({2}{x}+{3}\right)}{\left({x}-{1}\right)}={0}$

$\displaystyle{x}=-\frac{{3}}{{2}},{x}={1}$

Ignore the x=-3/2. The original problem states that x>0.

When x=1, y=2. The intersection point is (1,2).

Find the intersection point of $\displaystyle{y}=-{x}+{3}$ and $\displaystyle{y}=\frac{{1}}{{4}}{{x}}^{{2}}$ .

$\displaystyle\frac{{1}}{{4}}{{x}}^{{2}}=-{x}+{3}$

$\displaystyle{{x}}^{{2}}+{4}{x}-{12}={0}$

$\displaystyle{\left({x}+{6}\right)}{\left({x}-{2}\right)}={0}$

$\displaystyle{x}=-{6},{x}={2}$

Ignore the x=-6. The original problem states that x^0.

When x=2, y=1. The intersection point is (2, 1).

$\displaystyle{A}={\int_{{0}}^{{1}}}{\left({2}{{x}}^{{2}}-\frac{{1}}{{4}}{{x}}^{{2}}\right)}{\left.{d}{x}\right.}+{\int_{{1}}^{{2}}}{\left(-{x}+{3}-\frac{{1}}{{4}}{{x}}^{{2}}\right)}{\left.{d}{x}\right.}$

$\displaystyle={\int_{{0}}^{{1}}}{\left(\frac{{7}}{{4}}{{x}}^{{2}}\right)}{\left.{d}{x}\right.}-{\int_{{1}}^{{2}}}{\left(\frac{{1}}{{4}}{{x}}^{{2}}+{x}-{3}\right)}{\left.{d}{x}\right.}$

$\displaystyle={{\left[\frac{{7}}{{4}}\cdot\frac{{{x}}^{{3}}}{{3}}\right]}_{{0}}^{{1}}}-{{\left[\frac{{1}}{{4}}\cdot\frac{{{x}}^{{3}}}{{3}}+\frac{{{x}}^{{2}}}{{2}}-{3}{x}\right]}_{{1}}^{{2}}}$

$\displaystyle={{\left[\frac{{7}}{{12}}{{x}}^{{3}}\right]}_{{0}}^{{1}}}-{{\left[\frac{{1}}{{12}}{{x}}^{{3}}+\frac{{{x}}^{{2}}}{{2}}-{3}{x}\right]}_{{1}}^{{2}}}$

$\displaystyle={\left[\frac{{7}}{{12}}{{\left({1}\right)}}^{{3}}-{0}\right]}-{\left[{\left(\frac{{1}}{{12}}{{\left({2}\right)}}^{{3}}+\frac{{{\left({2}\right)}}^{{2}}}{{2}}-{3}{\left({2}\right)}\right)}-{\left(\frac{{1}}{{12}}{{\left({1}\right)}}^{{3}}+\frac{{{\left({1}\right)}}^{{2}}}{{2}}-{3}{\left({1}\right)}\right)}\right]}$