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

Wednesday, 26 February 2014

$\displaystyle{y}={{x}}^{{2}},{y}={4}{x}-{{x}}^{{2}}$ Sketch the region enclosed by the given curves and find its area.

You need to determine first the points of intersection between curves $\displaystyle{y}={{x}}^{{2}}$ and $\displaystyle{y}={4}{x}-{{x}}^{{2}}$ , by solving the equation, such that:

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

Factoring out 2x yields:

$\displaystyle{2}{x}{\left({x}-{2}\right)}={0}\Rightarrow{2}{x}={0}{\quad\text{or}\quad}{x}-{2}={0}$

Hence, the endpoints of integral are x = 0 and x = 2.

You need to decide...

You need to determine first the points of intersection between curves $\displaystyle{y}={{x}}^{{2}}$ and $\displaystyle{y}={4}{x}-{{x}}^{{2}}$ , by solving the equation, such that:

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

Factoring out 2x yields:

$\displaystyle{2}{x}{\left({x}-{2}\right)}={0}\Rightarrow{2}{x}={0}{\quad\text{or}\quad}{x}-{2}={0}$

Hence, the endpoints of integral are x = 0 and x = 2.

You need to decide what curve is greater than the other on the interval [0,2]. You need to notice that $\displaystyle{{x}}^{{2}}<{4}{x}-{{x}}^{{2}}$ on the interval [0,2], hence, you may evaluate the area of the region enclosed by the given curves, such that:

$\displaystyle{\int_{{a}}^{{b}}}{\left({f{{\left({x}\right)}}}-{g{{\left({x}\right)}}}\right)}{\left.{d}{x}\right.}$ , where $\displaystyle{f{{\left({x}\right)}}}>{g{{\left({x}\right)}}}$ for $\displaystyle{x}\in{\left[{a},{b}\right]}$

$\displaystyle{\int_{{0}}^{{2}}}{\left({4}{x}-{{x}}^{{2}}-{{x}}^{{2}}\right)}{\left.{d}{x}\right.}={\int_{{0}}^{{2}}}{\left({4}{x}-{2}{{x}}^{{2}}\right)}{\left.{d}{x}\right.}$

$\displaystyle{\int_{{0}}^{{2}}}{\left({4}{x}-{2}{{x}}^{{2}}\right)}{\left.{d}{x}\right.}={\int_{{0}}^{{2}}}{\left({4}{x}\right)}{\left.{d}{x}\right.}-{\int_{{0}}^{{2}}}{\left({2}{{x}}^{{2}}\right)}{\left.{d}{x}\right.}$

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

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

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

$\displaystyle{\int_{{0}}^{{2}}}{\left({4}{x}-{2}{{x}}^{{2}}\right)}{\left.{d}{x}\right.}={8}-\frac{{16}}{{3}}$

$\displaystyle{\int_{{0}}^{{2}}}{\left({4}{x}-{2}{{x}}^{{2}}\right)}{\left.{d}{x}\right.}=\frac{{{24}-{16}}}{{3}}$

$\displaystyle{\int_{{0}}^{{2}}}{\left({4}{x}-{2}{{x}}^{{2}}\right)}{\left.{d}{x}\right.}=\frac{{8}}{{3}}$

Hence, evaluating the area of the region enclosed by the given curves, yields $\displaystyle{\int_{{0}}^{{2}}}{\left({4}{x}-{2}{{x}}^{{2}}\right)}{\left.{d}{x}\right.}=\frac{{8}}{{3}}.$

The area of the region enclosed by the given curves is found between the red and orange curves, for $\displaystyle{x}\in{\left[{0},{2}\right]}$ .

Notes

Wednesday, 26 February 2014

$\displaystyle{y}={{x}}^{{2}},{y}={4}{x}-{{x}}^{{2}}$ Sketch the region enclosed by the given curves and find its area.

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Is there any personification in "The Tell-Tale Heart"?

Wednesday, 26 February 2014

Sketch the region enclosed by the given curves and find its area.

No comments:

Post a Comment

Is there any personification in &quot;The Tell-Tale Heart&quot;?

$\displaystyle{y}={{x}}^{{2}},{y}={4}{x}-{{x}}^{{2}}$ Sketch the region enclosed by the given curves and find its area.

Is there any personification in "The Tell-Tale Heart"?