If $\int_{0}^{1} (4 x^{3} - f (x)) f (x) d x = \frac{4}{7}$ then find the area of the region bounded by $y = f (x),$ X –axis and ordinates $x = 1 & x = 2.$

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answer is 8.

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Detailed Solution

$\int_{0}^{1} (4 x^{3} f (x) - f^{2} (x) + 4 x^{6} - 4 x^{6}) d x = \frac{4}{7}; \int_{0}^{a} 4 x^{6} d x - \int_{0}^{1} {(f (x) - 2 x^{3})}^{2} d x = \frac{4}{7}$

$\int_{0}^{1} {(f (x) - 2 x^{3})}^{2} d x = 0 ∴ f (x) = 2 x^{3}; \int_{0}^{2} f (x) d x = \frac{15}{2}$

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