Notes: Find the area bounded by y=x-2 and y=x^2-3

Friday, October 23, 2009

Find the area bounded by y=x-2 and y=x^2-3

Hello!

The graph of the first function, $\displaystyle{f{{\left({x}\right)}}}={x}-{2},$ is a straight line, the graph of the second function, $\displaystyle{g{{\left({x}\right)}}}={{x}}^{{2}}-{3},$ is a parabola branches up. These graphs have two intersections, $\displaystyle{\left({x}_{{1}},{y}_{{1}}\right)}$ and $\displaystyle{\left({x}_{{2}},{y}_{{2}}\right)},$ we'll find them. The figure whose area we have to compute lies between these points of intersection, and the first graph is over the second between $\displaystyle{x}_{{1}}$ and $\displaystyle{x}_{{2}}.$

Therefore the area $\displaystyle{A}$ is equal to $\displaystyle{\int_{{{x}_{{1}}}}^{{{x}_{{2}}}}}{\left({f{{\left({x}\right)}}}-{g{{\left({x}\right)}}}\right)}{\left.{d}{x}\right.}.$

To find $\displaystyle{x}_{{1}}$ and $\displaystyle{x}_{{2}}$ we have to solve the equation $\displaystyle{f{{\left({x}\right)}}}={g{{\left({x}\right)}}},$ i.e. $\displaystyle{x}-{2}={{x}}^{{2}}-{3},$ or $\displaystyle{{x}}^{{2}}-{x}-{1}={0}.$ It is a quadratic equation and $\displaystyle{x}_{{1}}=\frac{{{1}-\sqrt{{{5}}}}}{{2}},$ $\displaystyle{x}_{{2}}=\frac{{{1}+\sqrt{{{5}}}}}{{2}}.$

Thus $\displaystyle{A}={\int_{{{x}_{{1}}}}^{{{x}_{{2}}}}}{\left({f{{\left({x}\right)}}}-{g{{\left({x}\right)}}}\right)}{\left.{d}{x}\right.}={\int_{{{x}_{{1}}}}^{{{x}_{{2}}}}}{\left(-{{x}}^{{2}}+{x}+{1}\right)}{\left.{d}{x}\right.}=$

$\displaystyle={\left(-\frac{{1}}{{3}}{{x}}^{{3}}+\frac{{1}}{{2}}{{x}}^{{2}}+{x}\right)}{{\mid}_{{{x}_{{1}}}}^{{{x}_{{2}}}}}=-\frac{{1}}{{3}}{\left({{x}_{{2}}^{{3}}}-{{x}_{{1}}^{{3}}}\right)}+\frac{{1}}{{2}}{\left({{x}_{{2}}^{{2}}}-{{x}_{{1}}^{{2}}}\right)}+{\left({x}_{{2}}-{x}_{{1}}\right)}.$

We can simplify this slightly,

$\displaystyle{A}={\left({x}_{{2}}-{x}_{{1}}\right)}{\left(-\frac{{1}}{{3}}{\left({{x}_{{1}}^{{2}}}+{x}_{{1}}{x}_{{2}}+{{x}_{{2}}^{{2}}}\right)}+\frac{{1}}{{2}}{\left({x}_{{2}}+{x}_{{1}}\right)}+{1}\right)}=$

$\displaystyle=\sqrt{{{5}}}\cdot{\left(-\frac{{1}}{{3}}\cdot{\left({3}-{1}\right)}+\frac{{1}}{{2}}+{1}\right)}=\sqrt{{{5}}}\cdot{\left(\frac{{3}}{{2}}-\frac{{2}}{{3}}\right)}=\sqrt{{{5}}}\cdot\frac{{5}}{{6}}\approx{1.86}.$

This is the answer.

Notes

Friday, October 23, 2009

Find the area bounded by y=x-2 and y=x^2-3

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