Evaluate int_{-2}^1 f(x) dx, where f(x) = beg | vedclass

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(NONE) To evaluate the integral $\int_{-2}^1 f(x) dx$, we split the integral at $x=0$ because the definition of $f(x)$ changes at this point.
$\int_{

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$8$

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(NONE) To evaluate the integral $\int_{-2}^1 f(x) dx$, we split the integral at $x=0$ because the definition of $f(x)$ changes at this point.
$\int_{-2}^1 f(x) dx = \int_{-2}^0 (1-2x) dx + \int_0^1 (1+2x) dx$
Now, integrate each part separately:
$\int_{-2}^0 (1-2x) dx = [x - x^2]_{-2}^0 = (0 - 0) - (-2 - (-2)^2) = 0 - (-2 - 4) = 0 - (-6) = 6$
$\int_0^1 (1+2x) dx = [x + x^2]_0^1 = (1 + 1^2) - (0 + 0^2) = 2 - 0 = 2$
Adding these results together:
$\int_{-2}^1 f(x) dx = 6 + 2 = 8$

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

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