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Q.
The area bounded by the two branches of curve $(y-x)^{2}=x^{3}$
and the straight line $x = 1$ is
Application of Integrals
Solution:
$(y-x)^{2}=x^{3}$, where $x \,\ge\,0$
$\Rightarrow y-x=\pm x^{3 /2}$
or $ y=x+x^{3 /2} \left(1\right)$
$y=x-x^{3/ 2} \left(2\right)$
Function $\left(1\right)$ is an increasing function
Function $\left(2\right)$ meets $x$-axis, when $x-x^{3/ 2}=0$ or $x=0, 1$
Also, for $0<\,x<\,1, x-x^{3/ 2}>\,0$ and for $x >\,1, x-x^{3/ 2} <\,0$
When $x \rightarrow\infty, x-x^{3 /2} \rightarrow -\infty$
From these information, we can plot the graph as shown
Required area $=\int\limits_{0}^{1} \left[\left(x+x^{3/ 2}\right)-\left(x-x^{3/ 2}\right)\right]dx$
$=2 \int\limits_{0}^{1} x^{3 /2}dx$
$=2\left[\frac{x^{5 /2}}{5/ 2}\right]_{0}^{1} $
$=\frac{4}{5}$ sq. units
