int_{-5}^{5} |x+2| , dx is equal to | vedclass

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(A) We need to evaluate the integral $ I = \int_{-5}^{5} |x+2| \, dx $. Since $|x+2| = -(x+2)$ for $x < -2$ and $|x+2| = (x+2)$ for $x \ge -2$,we spli

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$ 29 $

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(A) We need to evaluate the integral $ I = \int_{-5}^{5} |x+2| \, dx $. Since $|x+2| = -(x+2)$ for $x $ I = \int_{-5}^{-2} -(x+2) \, dx + \int_{-2}^{5} (x+2) \, dx $ Evaluating the first part: $ -\left[ \frac{x^2}{2} + 2x \right]_{-5}^{-2} = -\left( (\frac{4}{2} - 4) - (\frac{25}{2} - 10) \right) = -\left( -2 - 2.5 \right) = 4.5 $ Evaluating the second part: $ \left[ \frac{x^2}{2} + 2x \right]_{-2}^{5} = \left( (\frac{25}{2} + 10) - (\frac{4}{2} - 4) \right) = (12.5 + 10) - (-2) = 22.5 + 2 = 24.5 $ Adding both parts: $ I = 4.5 + 24.5 = 29 $

$ \int_{-5}^{5} |x+2| \, dx $ is equal to

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