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# Find $\int \limits_2^{-1} f(x) dx\;$, given $\;\int \limits_{-1}^4 f(x) dx =4$ and$\;\int \limits_2^4 [3-f(x) ] dx=7$

$\begin {array} {1 1} (a)\;2 \\ (b)\;-4 \\ (c)\;-5 \\ (d)\;4 \end {array}$

$\int \limits_2 ^ 4 3 dx- \int \limits _2^4 f(x) dx =7$
$3(2) - \int \limits _{2}^{-1} f(x) dx - \int \limits_{-1}^4 f(x) dx =7$
$6- \int \limits _2 ^{-1} f(x)dx -\int \limits _{-1}^{4} f(x) dx =7$
$6- \int \limits _2 ^{-1} f(x)dx -4=7$
$\int \limits _2 ^{-1} f(x)dx =-5$
Hence c is the correct answer.