Evaluate the integral: int_0^2 [x] , dx + int | vedclass

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(D) We need to evaluate $I = \int_0^2 [x] \, dx + \int_0^2 |x-1| \, dx$. Step $1$: Evaluate $\int_0^2 [x] \, dx$. Since $[x] = 0$ for $0 \le x < 1$ an

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$2$

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(D) We need to evaluate $I = \int_0^2 [x] \, dx + \int_0^2 |x-1| \, dx$. Step $1$: Evaluate $\int_0^2 [x] \, dx$. Since $[x] = 0$ for $0 \le x $\int_0^2 [x] \, dx = \int_0^1 0 \, dx + \int_1^2 1 \, dx = 0 + [x]_1^2 = 2 - 1 = 1$. Step $2$: Evaluate $\int_0^2 |x-1| \, dx$. Since $|x-1| = -(x-1)$ for $0 \le x $\int_0^2 |x-1| \, dx = \int_0^1 (1-x) \, dx + \int_1^2 (x-1) \, dx$. $= [x - \frac{x^2}{2}]_0^1 + [\frac{x^2}{2} - x]_1^2 = (1 - \frac{1}{2}) - 0 + (2 - 2) - (\frac{1}{2} - 1) = \frac{1}{2} + \frac{1}{2} = 1$. Step $3$: Add the results. $I = 1 + 1 = 2$.

Evaluate the integral: $\int_0^2 [x] \, dx + \int_0^2 |x-1| \, dx$,where $[x]$ denotes the greatest integer function.

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