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(NONE OF THE ABOVE) To evaluate the integral $\int_{-2}^1 f(x) dx$,we split the integral at $x=0$ because the function definition changes at this poin

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(NONE OF THE ABOVE) To evaluate the integral $\int_{-2}^1 f(x) dx$,we split the integral at $x=0$ because the function definition changes at this point.$\int_{-2}^1 f(x) dx = \int_{-2}^0 f(x) dx + \int_0^1 f(x) dx$Substitute the given definitions of $f(x)$ for each interval:$= \int_{-2}^0 (1-2x) dx + \int_0^1 (1+2x) dx$Now,integrate each part:$= [x - x^2]_{-2}^0 + [x + x^2]_0^1$Evaluate the definite integrals:$= (0 - 0) - (-2 - (-2)^2) + (1 + 1^2) - (0 + 0^2)$$= 0 - (-2 - 4) + (1 + 1) - 0$$= -(-6) + 2 = 6 + 2 = 8$

Evaluate $\int_{-2}^1 f(x) dx$,where $f(x)= \begin{cases} 1-2x, & x \leq 0 \\ 1+2x, & x > 0 \end{cases}$

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