정적분 상수 결정

$\displaystyle \int _{0} ^{2} f \left( x \right) dx - \int _{0} ^{- 2} f \left( x \right) dx$$\displaystyle = \int _{0} ^{2} f \left( x \right) dx + \int _{- 2} ^{0} f \left( x \right) dx$ $\displaystyle = \int _{- 2} ^{2} f \left( x \right) dx = \frac{40}{3}$ $\displaystyle \int _{- 2} ^{2} \left( x ^{3} + x ^{2} + a \right) dx$$\displaystyle = 2 \int _{0} ^{2} \left( x ^{2} + a \right) dx$ $\displaystyle = 2 \left[ \begin{array}{l} \begin{matrix} \\ \end{matrix} \frac{1}{3} x ^{3} + ax \end{array} \right] _{0} ^{2}$ $\displaystyle = \frac{16}{3} + 4 a$ $\displaystyle \frac{16}{3} + 4 a = \frac{40}{3}$, $4 a = 8$ 따라서 $a = 2$