If int_{-1}^4 f(x) dx = 4 and int_2^4 (3 - f( | vedclass

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(A) We are given that $\int_{-1}^4 f(x) dx = 4$. Using the property of definite integrals,we can split the interval: $\int_{-1}^4 f(x) dx = \int_{-1}^

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$5$

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(A) We are given that $\int_{-1}^4 f(x) dx = 4$. Using the property of definite integrals,we can split the interval: $\int_{-1}^4 f(x) dx = \int_{-1}^2 f(x) dx + \int_2^4 f(x) dx = 4$. We are also given $\int_2^4 (3 - f(x)) dx = 7$. This can be written as: $\int_2^4 3 dx - \int_2^4 f(x) dx = 7$. Evaluating the first part: $[3x]_2^4 - \int_2^4 f(x) dx = 7 (3 	imes 4) - (3 	imes 2) - \int_2^4 f(x) dx = 7 12 - 6 - \int_2^4 f(x) dx = 7 6 - \int_2^4 f(x) dx = 7 \int_2^4 f(x) dx = 6 - 7 = -1$. Now,substitute $\int_2^4 f(x) dx = -1$ into the first equation: $\int_{-1}^2 f(x) dx + (-1) = 4 \int_{-1}^2 f(x) dx = 4 + 1 = 5$.

If $\int_{-1}^4 f(x) dx = 4$ and $\int_2^4 (3 - f(x)) dx = 7$,then $\int_{-1}^2 f(x) dx = $

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