## Thomas' Calculus 13th Edition

$$10\sqrt 3$$
\eqalign{ & \int_{ - \sqrt 3 }^{\sqrt 3 } {\left( {t + 1} \right)\left( {{t^2} + 4} \right)} dt \cr & {\text{use distributive property}} \cr & = \int_{ - \sqrt 3 }^{\sqrt 3 } {\left( {{t^3} + 4t + {t^2} + 4} \right)} dt \cr & {\text{integrate by using }}\int_a^b {{t^n}} dt = \left( {\frac{{{t^{n + 1}}}}{{n + 1}}} \right)_a^b \cr & = \left( {\frac{{{t^4}}}{4} + 2{t^2} + \frac{{{t^3}}}{3} + 4t} \right)_{ - \sqrt 3 }^{\sqrt 3 } \cr & {\text{Evaluating, we get:}} \cr & = \left( {\frac{{{{\left( {\sqrt 3 } \right)}^4}}}{4} + 2{{\left( {\sqrt 3 } \right)}^2} + \frac{{{{\left( {\sqrt 3 } \right)}^3}}}{3} + 4\left( {\sqrt 3 } \right)} \right) - \left( {\frac{{{{\left( { - \sqrt 3 } \right)}^4}}}{4} + 2{{\left( { - \sqrt 3 } \right)}^2} + \frac{{{{\left( { - \sqrt 3 } \right)}^3}}}{3} + 4\left( { - \sqrt 3 } \right)} \right) \cr & = \frac{9}{4} + 6 + \frac{{3\sqrt 3 }}{3} + 4\sqrt 3 - \frac{9}{4} - 6 + \frac{{3\sqrt 3 }}{3} + 4\sqrt 3 \cr & = \sqrt 3 + 4\sqrt 3 + \sqrt 3 + 4\sqrt 3 \cr & = 10\sqrt 3 \cr}