Notes: `int_0^(pi/2)(|sin(x) - cos(2x)|)dx` Evaluate the integral and interpret it as the area of a region. Sketch the region.

Friday, 27 September 2013

$\displaystyle{\int_{{0}}^{{\frac{\pi}{{2}}}}}{\left({\left|{\sin{{\left({x}\right)}}}-{\cos{{\left({2}{x}\right)}}}\right|}\right)}{\left.{d}{x}\right.}$ Evaluate the integral and interpret it as the area of a region. Sketch the region.

$\displaystyle{\int_{{0}}^{{\frac{\pi}{{2}}}}}$ (|sin(x)-cos(2x)|)dx

Since $\displaystyle{\sin{{\left({x}\right)}}}-{\cos{{\left({2}{x}\right)}}}\le{0},{\left[{0},\frac{\pi}{{6}}\right]}$

and $\displaystyle{\sin{{\left({x}\right)}}}-{\cos{{\left({2}{x}\right)}}}\ge{0},{\left[\frac{\pi}{{6}},\frac{\pi}{{2}}\right]}$

So, the integral can be split as,

$\displaystyle={\int_{{0}}^{{\frac{\pi}{{6}}}}}{\left(-{\left({\sin{{\left({x}\right)}}}-{\cos{{\left({2}{x}\right)}}}\right)}\right)}{\left.{d}{x}\right.}+{\int_{{\frac{\pi}{{6}}}}^{{\frac{\pi}{{2}}}}}{\left({\sin{{\left({x}\right)}}}-{\cos{{\left({2}{x}\right)}}}\right)}{\left.{d}{x}\right.}$

$\displaystyle={\int_{{0}}^{{\frac{\pi}{{6}}}}}{\left({\cos{{\left({2}{x}\right)}}}-{\sin{{\left({x}\right)}}}\right)}{\left.{d}{x}\right.}+{\int_{{\frac{\pi}{{6}}}}^{{\frac{\pi}{{2}}}}}{\left({\sin{{\left({x}\right)}}}-{\cos{{\left({2}{x}\right)}}}\right)}{\left.{d}{x}\right.}$

$\displaystyle={{\left[\frac{{1}}{{2}}{\sin{{\left({2}{x}\right)}}}+{\cos{{\left({x}\right)}}}\right]}_{{0}}^{{\frac{\pi}{{6}}}}}+{{\left[-{\cos{{\left({x}\right)}}}-\frac{{1}}{{2}}{\sin{{\left({2}{x}\right)}}}\right]}_{{\frac{\pi}{{6}}}}^{{\frac{\pi}{{2}}}}}$

$\displaystyle={\left(\frac{{1}}{{2}}{\sin{{\left(\frac{\pi}{{3}}\right)}}}+{\cos{{\left(\frac{\pi}{{6}}\right)}}}-{\left(\frac{{1}}{{2}}{\sin{{\left({0}\right)}}}+{\cos{{\left({0}\right)}}}\right)}+{\left(-{\cos{{\left(\frac{\pi}{{2}}\right)}}}-\frac{{1}}{{2}}{\sin{{\left(\pi\right)}}}\right)}-{\left(-{\cos{{\left(\frac{\pi}{{6}}\right)}}}-\frac{{1}}{{2}}{\sin{{\left(\frac{\pi}{{3}}\right)}}}\right)}\right.}$

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

$\displaystyle={\left(\frac{\sqrt{{{3}}}}{{4}}+\frac{\sqrt{{{3}}}}{{2}}-{1}+\frac{\sqrt{{{3}}}}{{2}}+\frac{\sqrt{{{3}}}}{{4}}\right)}$

$\displaystyle={\left(\frac{{3}}{{2}}\sqrt{{{3}}}-{1}\right)}$

$\displaystyle\approx{1.598}$

Graph is attached. Integral is the sum of the region from (0 to pi/6) and (pi/6 to pi/2).

$\displaystyle{\int_{{0}}^{{\frac{\pi}{{2}}}}}$ (|sin(x)-cos(2x)|)dx

Since $\displaystyle{\sin{{\left({x}\right)}}}-{\cos{{\left({2}{x}\right)}}}\le{0},{\left[{0},\frac{\pi}{{6}}\right]}$

and $\displaystyle{\sin{{\left({x}\right)}}}-{\cos{{\left({2}{x}\right)}}}\ge{0},{\left[\frac{\pi}{{6}},\frac{\pi}{{2}}\right]}$

So, the integral can be split as,

$\displaystyle={\left(\frac{\sqrt{{{3}}}}{{4}}+\frac{\sqrt{{{3}}}}{{2}}-{1}+\frac{\sqrt{{{3}}}}{{2}}+\frac{\sqrt{{{3}}}}{{4}}\right)}$

$\displaystyle={\left(\frac{{3}}{{2}}\sqrt{{{3}}}-{1}\right)}$

$\displaystyle\approx{1.598}$

Graph is attached. Integral is the sum of the region from (0 to pi/6) and (pi/6 to pi/2).

Notes

Friday, 27 September 2013

$\displaystyle{\int_{{0}}^{{\frac{\pi}{{2}}}}}{\left({\left|{\sin{{\left({x}\right)}}}-{\cos{{\left({2}{x}\right)}}}\right|}\right)}{\left.{d}{x}\right.}$ Evaluate the integral and interpret it as the area of a region. Sketch the region.

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

Friday, 27 September 2013

Evaluate the integral and interpret it as the area of a region. Sketch the region.

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

$\displaystyle{\int_{{0}}^{{\frac{\pi}{{2}}}}}{\left({\left|{\sin{{\left({x}\right)}}}-{\cos{{\left({2}{x}\right)}}}\right|}\right)}{\left.{d}{x}\right.}$ Evaluate the integral and interpret it as the area of a region. Sketch the region.

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