Let where , if the Area of the region bounded by the curves , axis, and is ; (where are coprime numbers) then

Question

Let $f (x) = \max {x^{2}, {(1 - x)}^{2}, 2 x (1 - x)}$ where $0 \leq x \leq 1$ , if the Area of the region bounded by the curves $y = f (x)$ , $x$ -axis, $x = 0$ and $x = 1$ is $\frac{p}{q}$ ; (where $p, q$ are coprime numbers) then $p + q =$ ____.

Smart Achievers · Accepted Answer

(3)

$The required Area A = \int_{0}^{1} f (x) d x$

$= \int_{0}^{1 / 3} {(1 - x)}^{2} d x + \int_{1 / 3}^{2 / 3} 2 x (1 - x) d x + \int_{2 / 3}^{1} x^{2} d x$

$= {[- \frac{1}{3} {(1 - x)}^{3}]}_{0}^{1 / 3} + {[x^{2} - \frac{2 x^{3}}{3}]}_{1 / 3}^{2 / 3} +$ ${[\frac{x^{3}}{3}]}_{\frac{2}{3}}^{1}$

$= \frac{17}{27}$

Solution

Q.
Let $f (x) = \max {x^{2}, {(1 - x)}^{2}, 2 x (1 - x)}$ where $0 \leq x \leq 1$ , if the Area of the region bounded by the curves $y = f (x)$ , $x$ -axis, $x = 0$ and $x = 1$ is $\frac{p}{q}$ ; (where $p, q$ are coprime numbers) then $p + q =$ ____.

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