## University Calculus: Early Transcendentals (3rd Edition)

$b^{2}-a^{2}$
Using equation (2), we obtain $\int^{b}_{a}2sds= 2\times[\frac{b^{2}}{2}-\frac{a^{2}}{2}]$ $=2\times\frac{b^{2}-a^{2}}{2}=b^{2}-a^{2}$