The area contained between the curve $x y = a^{2}$ , the vertical line x = a, x = 4a(a > 0) and x-axis is

a^{2} l o g 2

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2 a^{2} l o g 2

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a l o g 2

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2 a l o g 2

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Solution

The correct option is B $2 a^{2} l o g 2$
Area = $\int_{a}^{4 a} \frac{a^{2}}{x} d x = [a^{2} l o g x]_{a}^{4 a} = a^{2} [l o g 4 a - l o g a] = a^{2} l o g [\frac{4 a}{a}] = a^{2} l o g 4 = 2 a^{2} l o g 2$

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