Set $up$ the estimation $of$ the integral using the trapezoid rule and four

subintervals. You need not solve $it.$

${\displaystyle \underset{1}{\overset{3}{}}}\sqrt[\phantom{2}]{1+x^2}dx$

$\sqrt[\phantom{2}]{2}+\sqrt[\phantom{2}]{13}+2\sqrt[\phantom{2}]{5}+\sqrt[\phantom{2}]{29}+2\sqrt[\phantom{2}]{10}$

$\frac{1}{4}(\sqrt[\phantom{2}]{2}+\sqrt[\phantom{2}]{13}+2\sqrt[\phantom{2}]{5}+\sqrt[\phantom{2}]{29}+\sqrt[\phantom{2}]{10})$

$\sqrt[\phantom{2}]{2}+\sqrt[\phantom{2}]{13}+\sqrt[\phantom{2}]{5}+\sqrt[\phantom{2}]{29}$

$2(\sqrt[\phantom{2}]{2}+\sqrt[\phantom{2}]{13}+2\sqrt[\phantom{2}]{5}+\sqrt[\phantom{2}]{29}+\sqrt[\phantom{2}]{10})$

$\frac{1}{4}(\sqrt[\phantom{2}]{2}+\sqrt[\phantom{2}]{13}+2\sqrt[\phantom{2}]{5}+\sqrt[\phantom{2}]{29})$