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

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(C) Given that $\int_{2}^{4} (3 - f(x)) \, dx = 7$.This can be written as $\int_{2}^{4} 3 \, dx - \int_{2}^{4} f(x) \, dx = 7$.Evaluating the first in

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$-5$

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(C) Given that $\int_{2}^{4} (3 - f(x)) \, dx = 7$.This can be written as $\int_{2}^{4} 3 \, dx - \int_{2}^{4} f(x) \, dx = 7$.Evaluating the first integral: $[3x]_{2}^{4} = 3(4 - 2) = 6$.So,$6 - \int_{2}^{4} f(x) \, dx = 7$,which implies $\int_{2}^{4} f(x) \, dx = -1$.We know that $\int_{-1}^{4} f(x) \, dx = \int_{-1}^{2} f(x) \, dx + \int_{2}^{4} f(x) \, dx$.Substituting the known values: $4 = \int_{-1}^{2} f(x) \, dx + (-1)$.Therefore,$\int_{-1}^{2} f(x) \, dx = 4 + 1 = 5$.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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