Question

Find the area bounded between the curves Y=xsquare ,Y=xcube

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Answer 1

We need to find area bounded between the curves,

• $y=x^2$

and,

• $y=x^3$

Assume,

$\longrightarrow x^3\leq x^2$

$\longrightarrow x^3-x^2\leq0$

$\longrightarrow x^2(x-1)\leq0$

Performing wavy curve method,

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\thicklines\put(0,0){\line(1,0){60}}\multiput(20,0)(20,0){2}{\circle*{1}}\put(19,-5){ 0 }\put(39,-5){ 1 }\qbezier(40,0)(50,20)(60,10)\qbezier(30,-10)(35,-10)(40,0)\qbezier(25,-5)(27.5,-10)(30,-10)\qbezier(20,0)(22.5,-0)(25,-5)\qbezier(0,-10)(10,0)(20,0)\end{picture}$

Then we get,

$\longrightarrow x\in\left(-\infty,\ 1]$

But since $x^2>0$ and $x^3<0$ for $x<0,$ no area is bounded by the two curves for $x<0$ as they diverge there.

So we consider,

$\longrightarrow x\in[0,\ 1]$

Hence the area bounded will be,

$\displaystyle\longrightarrow A=\int\limits_0^1(x^2-x^3)\ dx$

$\displaystyle\longrightarrow A=\dfrac{1}{3}\left[x^3\right]_0^1-\dfrac{1}{4}\left[x^4\right]_0^1$

$\displaystyle\longrightarrow A=\dfrac{1}{3}-\dfrac{1}{4}$

$\displaystyle\longrightarrow\underline{\underline{A=\dfrac{1}{12}}}$

Answer 2

☆ We have to find the area bounded between the curves

$\tt \:y \: = \: {x}^{2} \: and \: y \: = \: {x}^{3}$

$\tt \: \: \: \: \: y \: = \: {x}^{2} - - - (1)$

$\tt \: \: \: \: \: \: y \: = \: {x}^{3} - - - (2)$

$\large\underline{\bold{❥︎Step :- 1 }}$

☆ Point of intersection of (1) and (2)

$\tt \: {x}^{2} = {x}^{2}$

$\tt \implies \: {x}^{3} \: - \: {x}^{2} = 0$

$\tt \implies \: {x}^{2} (x - 1) = 0$

$\tt \implies \: x \: = \: 0 \: or \: x \: = \: 1$

➢ So, point of intersection of curve (1) and (2) are given below

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf 0 \\ \\ \sf 1 & \sf 1 \end{array}} \\ \end{gathered}$

$\large\underline{\bold{❥︎Step :- 2 }}$

Curve sketching :-

Please check the attachment for area.

$\tt \:Blue \: indicates \: y \: = \: {x}^{3} \\ \tt \: \:\: Red \: indicates \: y \: = \: {x}^{2} \: \: \: \:$

$\large\underline{\bold{❥︎Step :- 3 }}$

$\tt \:Required \: Area \: = \int_0^1x^2dx \: - \: \int_0^1x^3dx$

$\tt \implies \: = \bigg[ \dfrac{ {x}^{3} }{3} \bigg]_0^1 \: - \: \bigg[ \dfrac{ {x}^{4} }{4} \bigg]_0^1$

$\tt \implies \: = \dfrac{1}{3} - \dfrac{1}{4}$

$\tt \implies \: = \: \dfrac{1}{12} \: sq. \: units$