`int_1^2((v^3 + 3v^6)/(v^4))dv` Evaluate the integral.

$\displaystyle{\int_{{1}}^{{2}}}{\left(\frac{{{{v}}^{{3}}+{3}{{v}}^{{6}}}}{{{v}}^{{4}}}\right)}{d}{v}$

simplify the integrand and apply the sum rule,

$\displaystyle={\int_{{1}}^{{2}}}{\left(\frac{{{v}}^{{3}}}{{{v}}^{{4}}}+\frac{{{3}{{v}}^{{6}}}}{{{v}}^{{4}}}\right)}{d}{v}$

$\displaystyle={\int_{{1}}^{{2}}}{\left(\frac{{1}}{{v}}+{3}{{v}}^{{2}}\right)}{d}{v}$

using the following common integrals

$\displaystyle\int\frac{{1}}{{x}}{\left.{d}{x}\right.}={\ln{{\left({x}\right)}}}$ and $\displaystyle\int{{x}}^{{n}}=\frac{{{{x}}^{{{n}+{1}}}}}{{{n}+{1}}}$

$\displaystyle={\int_{{1}}^{{2}}}{\left(\frac{{1}}{{v}}\right)}{d}{v}+{3}{\int_{{1}}^{{2}}}{{v}}^{{2}}{d}{v}$

$\displaystyle={{\left[{\ln{{\left({v}\right)}}}\right]}_{{1}}^{{2}}}+{3}{{\left[\frac{{{v}}^{{3}}}{{3}}\right]}_{{1}}^{{2}}}$

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

$\displaystyle={\ln{{\left({2}\right)}}}+{3}{\left(\frac{{{8}-{1}}}{{3}}\right)}$

$\displaystyle={\ln{{\left({2}\right)}}}+{7}$

Spooky Notes

Tuesday, November 20, 2012

$\displaystyle{\int_{{1}}^{{2}}}{\left(\frac{{{{v}}^{{3}}+{3}{{v}}^{{6}}}}{{{{v}}^{{4}}}}\right)}{d}{v}$ Evaluate the integral.

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