int_{-2}^{1} [x+1] , dx = (Where [x] is the g | vedclass

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(B) Let $I = \int_{-2}^{1} [x+1] \, dx$. Using the property $[x+n] = [x] + n$ for any integer $n$,we have $[x+1] = [x] + 1$. So,$I = \int_{-2}^{1} ([x

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(B) Let $I = \int_{-2}^{1} [x+1] \, dx$. Using the property $[x+n] = [x] + n$ for any integer $n$,we have $[x+1] = [x] + 1$. So,$I = \int_{-2}^{1} ([x] + 1) \, dx = \int_{-2}^{1} [x] \, dx + \int_{-2}^{1} 1 \, dx$. Evaluating the second integral: $\int_{-2}^{1} 1 \, dx = [x]_{-2}^{1} = 1 - (-2) = 3$. Evaluating the first integral: $\int_{-2}^{1} [x] \, dx = \int_{-2}^{-1} -2 \, dx + \int_{-1}^{0} -1 \, dx + \int_{0}^{1} 0 \, dx$. $= -2[x]_{-2}^{-1} - 1[x]_{-1}^{0} + 0 = -2(-1 - (-2)) - 1(0 - (-1)) = -2(1) - 1(1) = -2 - 1 = -3$. Therefore,$I = -3 + 3 = 0$.

$\int_{-2}^{1} [x+1] \, dx =$ (Where $[x]$ is the greatest integer function not greater than $x$)

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