Question

Area Bounded between y = max(x², +(1-x)², 2x(1-x)), x=0, x=1 and x-axis.

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Answer 1

The area is bounded between $x=0$ and $x=1,$ so the limits of the integral to be performed should be between 0 and 1.

We need to find area under $y=\max\{x^2,\ (1-x)^2,\ 2x(1-x)\}$ between 0 and 1.

Assume,

$\longrightarrow x^2>(1-x)^2$

$\longrightarrow x^2>1-2x+x^2$

$\longrightarrow2x-1>0$

$\longrightarrow x\in\left(\dfrac{1}{2},\ \infty\right)$

This implies,

• $x\in\left[0,\ \dfrac{1}{2}\right]\quad\implies\quad (1-x)^2\geq x^2$

• $x\in\left[\dfrac{1}{2},\ 1\right]\quad\implies\quad x^2\geq(1-x)^2$

Assume,

$\longrightarrow(1-x)^2>2x(1-x)$

$\longrightarrow1-2x+x^2>2x-2x^2$

$\longrightarrow3x^2-4x+1>0$

$\longrightarrow(3x-1)(x-1)>0$

$\longrightarrow x\in\left(-\infty,\ \dfrac{1}{3}\right)\cup\bigg(1,\ \infty\bigg)$

This implies,

• $x\in\left[0,\ \dfrac{1}{3}\right]\quad\implies\quad(1-x)^2\geq 2x(1-x)$

• $x\in\left[\dfrac{1}{3},\ 1\right]\quad\implies\quad2x(1-x)\geq(1-x)^2$

Assume,

$\longrightarrow2x(1-x)>x^2$

$\longrightarrow2x-2x^2>x^2$

$\longrightarrow3x^2-2x<0$

$\longrightarrow x(3x-2)<0$

$\longrightarrow x\in\left(0,\ \dfrac{2}{3}\right)$

This implies,

• $x\in\left[0,\ \dfrac{2}{3}\right]\quad\implies\quad2x(1-x)\geq x^2$

• $x\in\left[\dfrac{2}{3},\ 1\right]\quad\implies\quad x^2\geq 2x(1-x)$

Analysing each, we get the following.

• $x\in\left[0,\ \dfrac{1}{3}\right]\quad\implies\quad y=(1-x)^2$

• $x\in\left[\dfrac{1}{3},\ \dfrac{1}{2}\right]\quad\implies\quad y=2x(1-x)$

• $x\in\left[\dfrac{1}{2},\ \dfrac{2}{3}\right]\quad\implies\quad y=2x(1-x)$

• $x\in\left[\dfrac{2}{3},\ 1\right]\quad\implies\quad y=x^2$

Hence the area bounded will be given by,

$\displaystyle\longrightarrow A=\int\limits_0^{\frac{1}{3}}(1-x)^2\ dx+\int\limits_{\frac{1}{3}}^{\frac{2}{3}}2x(1-x)\ dx+\int\limits_{\frac{2}{3}}^1x^2\ dx$

$\displaystyle\longrightarrow A=\int\limits_0^{\frac{1}{3}}\left(1-2x+x^2\right)\ dx+\int\limits_{\frac{1}{3}}^{\frac{2}{3}}\left(2x-2x^2\right)\ dx+\int\limits_{\frac{2}{3}}^1x^2\ dx$

$\displaystyle\longrightarrow A=\big[x\big]_0^{\frac{1}{3}}-\left[x^2\right]_0^{\frac{1}{3}}+\dfrac{1}{3}\left[x^3\right]_0^{\frac{1}{3}}+\left[x^2\right]_{\frac{1}{3}}^{\frac{2}{3}}-\dfrac{2}{3}\left[x^3\right]_{\frac{1}{3}}^{\frac{2}{3}}+\dfrac{1}{3}\left[x^3\right]_{\frac{2}{3}}^1$

$\displaystyle\longrightarrow A=\dfrac{1}{3}-\dfrac{1}{9}+\dfrac{1}{3}\times\dfrac{1}{27}+\dfrac{1}{3}-\dfrac{2}{3}\times\dfrac{7}{27}+\dfrac{1}{3}\times\dfrac{19}{27}$

$\displaystyle\longrightarrow\underline{\underline{A=\dfrac{17}{27}}}$