## Thomas' Calculus 13th Edition

$c$=$\sqrt {21}$
$\int^c_3\frac{1}{6}xdx$ =$\frac{1}{12}c^2-\frac{3}{4}$ solving $\frac{1}{12}c^2-\frac{3}{4}$ =$1$ we find $c$=$\sqrt {21}$