If int_{-1}^{4} f(x) dx = 4 and int_{2}^{4} ( | vedclass

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(C) Given $\int_{2}^{4} (3 - f(x)) dx = 7$. Expanding the integral: $\int_{2}^{4} 3 dx - \int_{2}^{4} f(x) dx = 7$. Since $\int_{2}^{4} 3 dx = [3x]_{2

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$-5$

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(C) Given $\int_{2}^{4} (3 - f(x)) dx = 7$. Expanding the integral: $\int_{2}^{4} 3 dx - \int_{2}^{4} f(x) dx = 7$. Since $\int_{2}^{4} 3 dx = [3x]_{2}^{4} = 3(4 - 2) = 6$,we have $6 - \int_{2}^{4} f(x) dx = 7$. Therefore,$\int_{2}^{4} f(x) dx = 6 - 7 = -1$. We need to find $\int_{2}^{-1} f(x) dx$. Using the property $\int_{a}^{b} f(x) dx = -\int_{b}^{a} f(x) dx$,we have $\int_{2}^{-1} f(x) dx = -\int_{-1}^{2} f(x) dx$. Using the property $\int_{a}^{c} f(x) dx = \int_{a}^{b} f(x) dx + \int_{b}^{c} f(x) dx$,we know $\int_{-1}^{4} f(x) dx = \int_{-1}^{2} f(x) dx + \int_{2}^{4} f(x) dx$. Substituting the known values: $4 = \int_{-1}^{2} f(x) dx + (-1)$. Thus,$\int_{-1}^{2} f(x) dx = 4 + 1 = 5$. Finally,$\int_{2}^{-1} f(x) dx = -\int_{-1}^{2} f(x) dx = -5$.

If $\int_{-1}^{4} f(x) dx = 4$ and $\int_{2}^{4} (3 - f(x)) dx = 7$,then the value of $\int_{2}^{-1} f(x) dx$ is

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