Evaluate the definite integral int_{-2}^{2.24 | vedclass

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(D) The integral is $\int_{-2}^{2.24} [x] \, dx$. We split the interval based on the jumps of the greatest integer function $[x]$:$\int_{-2}^{2.24} [x

Vedclass · Accepted Answer

$-1.52$

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(D) The integral is $\int_{-2}^{2.24} [x] \, dx$. We split the interval based on the jumps of the greatest integer function $[x]$:$\int_{-2}^{2.24} [x] \, dx = \int_{-2}^{-1} -2 \, dx + \int_{-1}^{0} -1 \, dx + \int_{0}^{1} 0 \, dx + \int_{1}^{2} 1 \, dx + \int_{2}^{2.24} 2 \, dx$Evaluating each part:$\int_{-2}^{-1} -2 \, dx = -2[-1 - (-2)] = -2(1) = -2$$\int_{-1}^{0} -1 \, dx = -1[0 - (-1)] = -1(1) = -1$$\int_{0}^{1} 0 \, dx = 0$$\int_{1}^{2} 1 \, dx = 1[2 - 1] = 1$$\int_{2}^{2.24} 2 \, dx = 2[2.24 - 2] = 2(0.24) = 0.48$Summing these values: $-2 - 1 + 0 + 1 + 0.48 = -2 + 0.48 = -1.52$.

Evaluate the definite integral $\int_{-2}^{2.24} [x] \, dx$,where $[x]$ denotes the greatest integer function.

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