`int_1^4 1/(xsqrt(16x^2-5)) dx` Evaluate the definite integral

Saturday, May 16, 2009

$\displaystyle{\int_{{1}}^{{4}}}\frac{{1}}{{{x}\sqrt{{{16}{{x}}^{{2}}-{5}}}}}{\left.{d}{x}\right.}$ Evaluate the definite integral

Make the substitution $\displaystyle{u}=\sqrt{{{16}{{x}}^{{2}}-{5}}},$ then $\displaystyle{16}{{x}}^{{2}}={{u}}^{{2}}+{5},$

$\displaystyle{d}{u}=\frac{{{32}{x}}}{{{2}\sqrt{{{16}{{x}}^{{2}}-{5}}}}}{\left.{d}{x}\right.}=\frac{{{16}{x}{\left.{d}{x}\right.}}}{\sqrt{{{16}{{x}}^{{2}}-{5}}}},$

or $\displaystyle\frac{{\left.{d}{x}}}{\sqrt{{{16}{{x}}^{{2}}-{5}}}}=\frac{{{d}{u}}}{{{16}{x}}}.$

The limits of integration for $\displaystyle{u}$ are from $\displaystyle\sqrt{{{11}}}$ to $\displaystyle\sqrt{{{251}}}.$

The indefinite integral becomes $\displaystyle\int\frac{{{d}{u}}}{{{16}{{x}}^{{2}}}}=\int\frac{{{d}{u}}}{{{{u}}^{{2}}+{5}}},$

which is equal to $\displaystyle\frac{{1}}{\sqrt{{{5}}}}{\arctan{{\left(\frac{{u}}{\sqrt{{{5}}}}\right)}}}+{C}.$

This way the definite integral is

$\displaystyle\frac{{1}}{\sqrt{{{5}}}}{\left({\arctan{{\left(\sqrt{{\frac{{251}}{{5}}}}\right)}}}-{\arctan{{\left(\sqrt{{\frac{{11}}{{5}}}}\right)}}}\right)}\approx{0.2026}.$

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Saturday, May 16, 2009

$\displaystyle{\int_{{1}}^{{4}}}\frac{{1}}{{{x}\sqrt{{{16}{{x}}^{{2}}-{5}}}}}{\left.{d}{x}\right.}$ Evaluate the definite integral

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Thomas Jefferson's election in 1800 is sometimes called the Revolution of 1800. Why could it be described in this way?

Saturday, May 16, 2009

Evaluate the definite integral

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Thomas Jefferson&#39;s election in 1800 is sometimes called the Revolution of 1800. Why could it be described in this way?

$\displaystyle{\int_{{1}}^{{4}}}\frac{{1}}{{{x}\sqrt{{{16}{{x}}^{{2}}-{5}}}}}{\left.{d}{x}\right.}$ Evaluate the definite integral

Thomas Jefferson's election in 1800 is sometimes called the Revolution of 1800. Why could it be described in this way?