Blog: `int_-1^1(|3^x - 2^x|)dx` Evaluate the integral and interpret it as the area of a region. Sketch the region.

Saturday, July 25, 2015

$\displaystyle{\int_{{-{{1}}}}^{{1}}}{\left({\left|{{3}}^{{x}}-{{2}}^{{x}}\right|}\right)}{\left.{d}{x}\right.}$ Evaluate the integral and interpret it as the area of a region. Sketch the region.

Since $\displaystyle{{3}}^{{x}}>{{2}}^{{x}}$ for $\displaystyle{x}>{0}$ and $\displaystyle{{3}}^{{x}}<{{2}}^{{x}}$ for $\displaystyle{x}<{0}$ we have

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

Now we apply formula $\displaystyle\int{{a}}^{{x}}=\frac{{{a}}^{{x}}}{{\ln{{a}}}}+{C}.$

$\displaystyle{\left(\frac{{{2}}^{{x}}}{{\ln{{2}}}}-\frac{{{3}}^{{x}}}{{\ln{{3}}}}\right)}{{\left|_{{-{{1}}}}^{{0}}+{\left(\frac{{{3}}^{{x}}}{{\ln{{3}}}}-\frac{{{2}}^{{x}}}{{\ln{{2}}}}\right)}\right|}_{{0}}^{{1}}}=$

$\displaystyle\frac{{1}}{{\ln{{2}}}}-\frac{{1}}{{\ln{{3}}}}-\frac{{1}}{{{2}{\ln{{2}}}}}+\frac{{1}}{{{3}{\ln{{3}}}}}+\frac{{3}}{{\ln{{3}}}}-\frac{{2}}{{\ln{{2}}}}-\frac{{1}}{{\ln{{3}}}}+\frac{{1}}{{\ln{{2}}}}=$

$\displaystyle\frac{{-{3}{\ln{{3}}}+{8}{\ln{{2}}}}}{{{6}{\ln{{2}}}{\ln{{3}}}}}=$

$\displaystyle\frac{{\ln{{\left(\frac{{256}}{{27}}\right)}}}}{{{\ln{{3}}}{\ln{{64}}}}}\approx{0.4923}$

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Saturday, July 25, 2015

$\displaystyle{\int_{{-{{1}}}}^{{1}}}{\left({\left|{{3}}^{{x}}-{{2}}^{{x}}\right|}\right)}{\left.{d}{x}\right.}$ Evaluate the integral and interpret it as the area of a region. Sketch the region.

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