Area bounded by the curve xy-3x-2y-10=0, x-axis and the lines x=3, x=4 is

16 log 2 - 13

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16 log 2 - 3

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16 log 2 + 3

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None of these

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Solution

The correct option is C 16 log 2 + 3
Given curve is $y (x - 2) = 3 x + 10 \Rightarrow y = \frac{3 x + 10}{x - 2}$
Required area is $\int_{3}^{4} y d x = \int_{3}^{4} \frac{3 x + 10}{x - 2} d x$
$= [3 x + 16 l o g (x - 2)]_{3}^{4} = 3 + 16 l o g 2$ = sq. unit.

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