The area enclosed by $y = 3 x - 5, x = 3$ and $x = 5$ is

12

sq units

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13

sq units

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13 \frac{1}{2}

sq units

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14

sq units

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Solution

The correct option is D $14$ sq units
Area $= \int_{3}^{5} (3 x - 5) d x$
$= \frac{3}{2} (x^{2})_{3}^{5} - 5 (x)_{3}^{5} = \frac{3}{2} [25 - 9] - 5 (5 - 3)$
$= \frac{3}{2} 16 - 5 (2) = 24 - 10 = 14$

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