IMathAS-libretexts

To find the arclength of

\displaystyle {y}={2}{x}^{{2}}-{3}

from

\displaystyle {x}={0}

\displaystyle {x}={5}

, which is the correct integral?

$\displaystyle {L}={\int_{{0}}^{{5}}}\sqrt{{{1}+{16}{x}^{{2}}}}{\left.{d}{x}\right.}$
$\displaystyle {L}={\int_{{0}}^{{5}}}\sqrt{{{1}+{16}{x}}}{\left.{d}{x}\right.}$
$\displaystyle {L}={\int_{{0}}^{{5}}}\sqrt{{{1}+{4}{x}}}{\left.{d}{x}\right.}$
$\displaystyle {L}={\int_{{0}}^{{5}}}\sqrt{{{1}+{4}{x}^{{2}}}}{\left.{d}{x}\right.}$

Now that you have the correct integral, use the substitution

\displaystyle {u}={4}{x}

to re-write the integral in terms of

\displaystyle {u}

. Which is the correct result?

$\displaystyle {L}=\frac{{1}}{{4}}{\int_{{0}}^{{20}}}\sqrt{{{1}+{u}}}{d}{u}$
$\displaystyle {L}=\frac{{1}}{{4}}{\int_{{0}}^{{20}}}\sqrt{{{1}+{2}{u}^{{2}}}}{d}{u}$
$\displaystyle {L}=\frac{{1}}{{4}}{\int_{{0}}^{{20}}}\sqrt{{{1}+{4}{u}}}{d}{u}$
$\displaystyle {L}=\frac{{1}}{{4}}{\int_{{0}}^{{20}}}\sqrt{{{1}+{u}^{{2}}}}{d}{u}$

What is the value of the integral? Hint: Use a table or technology. You may enter a decimal approximation for your solution.