Discussion
\(\int_{-2}^{2}|x + 3|\,dx =\)
Answer: E
The value is 12.
On [-2, 2], x + 3 is always positive (between 1 and 5), so \(|x + 3| = x + 3\).
\[ \begin{aligned} \int_{-2}^{2}(x + 3)\,dx &= \left[\frac{x^{2}}{2} + 3x\right]_{-2}^{2} \\ &= 8 - (-4) = 12 \end{aligned} \]
The x term integrates to zero over the symmetric interval, leaving \(3 \times 4 = 12\). 8 keeps only the upper limit and 16 adds instead of subtracting at the lower limit.
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