allery$\int_{-2}^{2} (x^3 + x^2)\,dx$ equals:
A$0$
B$8/3$
C$16/3$
D$32/3$
Answer & Solution
Correct answer: C. $16/3$
**Split:** $\int_{-2}^2 x^3 dx + \int_{-2}^2 x^2 dx$.
First integral = 0 ($x^3$ odd).
Second = $2\int_0^2 x^2\,dx$ ($x^2$ even) $= 2 \cdot 8/3 = 16/3$.
**Total:** $0 + 16/3 = 16/3$.
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