int_{-4}^{4} |x + 2| , dx = | vedclass

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(C) We evaluate the definite integral $\int_{-4}^{4} |x + 2| \, dx$ by splitting the interval at $x = -2$ where the expression inside the absolute val

Vedclass · Accepted Answer

$20$

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(C) We evaluate the definite integral $\int_{-4}^{4} |x + 2| \, dx$ by splitting the interval at $x = -2$ where the expression inside the absolute value changes sign.Since $|x + 2| = -(x + 2)$ for $x $\int_{-4}^{4} |x + 2| \, dx = \int_{-4}^{-2} -(x + 2) \, dx + \int_{-2}^{4} (x + 2) \, dx$Evaluating the first integral:$\int_{-4}^{-2} -(x + 2) \, dx = -\left[ \frac{x^2}{2} + 2x \right]_{-4}^{-2} = -\left[ (\frac{4}{2} - 4) - (\frac{16}{2} - 8) \right] = -[(-2) - (0)] = 2$Evaluating the second integral:$\int_{-2}^{4} (x + 2) \, dx = \left[ \frac{x^2}{2} + 2x \right]_{-2}^{4} = \left[ (\frac{16}{2} + 8) - (\frac{4}{2} - 4) \right] = [(8 + 8) - (-2)] = 16 + 2 = 18$Adding the two results:$2 + 18 = 20$Thus,the correct option is $C$.

$\int_{-4}^{4} |x + 2| \, dx = $

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