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Mathematics - Area Under Curves Question with Solution | TestHub

MathematicsArea Under CurvesArea bounded by two curvesEasy2 minQB

MathematicsEasysingle choice

Passage / Comprehension

Let $f (x)$ be a differentiable function satisfying $(x - y) f (x + y) - (x + y) f (x - y) = 4 x y (x^{2} - y^{2})$ for all $x, y \in R$ . If $f (1) = 1$ then

The area of the region bounded by the curves $y = f (x) and y = x^{2} is$

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Answer:

Solution:

Dividing by $(x^{2} - y^{2})$ with $g (x) = \frac{f ( x )}{x} : g (x + y) - g (x - y) = 4 x y$ . Put $u = x + y, v = x - y$ : $4 x y = u^{2} - v^{2}$ , so $g (u) - u^{2} = g (v) - v^{2} = c$ , a constant. Hence $f (x) = x^{3} + c x$ , and $f (1) = 1$ gives $c = 0$ , so $f (x) = x^{3}$ . $x^{3} = x^{2}$ at $x = 0, 1$ . Area $= \int_{0}^{1} (x^{2} - x^{3}) d x = \frac{1}{3} - \frac{1}{4} = \frac{1}{12}$ .

Stream:JEESubject:MathematicsTopic:Area Under CurvesSubtopic:Area bounded by two curves

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