Explanation
Given that,
The curve and the coordinate of any point on its equal to the cube of
that ordinates.
Let (h, k) be point on it is equal to the cube of that ordinate.
Given the area of the region bounded by the co-ordinate axis in above figure.
We know that the area of rectangle is
$$Area=length\times breath$$
But Given that ,
Ordinate of point $$=k$$
$$h\times k={{k}^{3}}\,$$ (given function )
$$h={{k}^{2}}$$
We have,
$${{x}^{2}}+{{y}^{2}}=1$$
$${{y}_{1}}=\sqrt{1-{{x}^{2}}}$$ ……. (1)
$$\left| y \right|=1-{{x}^{2}}$$
$$y=1-{{x}^{2}}$$ ……. (2)
So, area of first quadrant
$$=\int_{0}^{1}{\left( {{y}_{1}}-{{y}_{2}} \right)}dx$$
$$ =\int_{0}^{1}{\left( \sqrt{1-{{x}^{2}}}-\left( 1-{{x}^{2}} \right) \right)}dx $$
$$ =\int_{0}^{1}{\left( \sqrt{1-{{x}^{2}}} \right)}dx-\int_{0}^{1}{\left( 1-{{x}^{2}} \right)}dx $$
We know that
$$\int{\sqrt{{{a}^{2}}-{{x}^{2}}}dx}=\dfrac{1}{2}a\sqrt{{{a}^{2}}-{{x}^{2}}}+\dfrac{1}{2}{{a}^{2}}{{\sin }^{-1}}\left( \dfrac{x}{a} \right)+C$$
Therefore,
$$ =\left[ \dfrac{1}{2}\sqrt{1-{{x}^{2}}}+\dfrac{1}{2}{{\sin }^{-1}}\left( x \right) \right]_{0}^{1}-\left( x-\dfrac{{{x}^{3}}}{3} \right)_{0}^{1} $$
$$ =\left[ \left( \dfrac{1}{2}\sqrt{1-1}+\dfrac{1}{2}{{\sin }^{-1}}\left( 1 \right) \right)-\left( \dfrac{1}{2}\sqrt{1-0}+\dfrac{1}{2}{{\sin }^{-1}}\left( 0 \right) \right) \right]-\left[ \left( 1-\dfrac{1}{3} \right)-0 \right] $$
$$ =\left[ \left( 0+\dfrac{1}{2}{{\sin }^{-1}}\left( \sin \dfrac{\pi }{2} \right) \right)-\left( \dfrac{1}{2}+\dfrac{1}{2}{{\sin }^{-1}}\left( \sin 0 \right) \right) \right]-\dfrac{2}{3} $$
$$ =\dfrac{\pi }{4}-\dfrac{1}{2}-\dfrac{2}{3} $$
$$ =\dfrac{\pi }{4}-\left( \dfrac{3+4}{6} \right) $$
$$ =\dfrac{\pi }{4}-\dfrac{7}{6} $$
$$ =\dfrac{3\pi -14}{12} $$
Area of all quadrants
$$ =4\times \left( \dfrac{3\pi-14}{12} \right) $$
$$ =\dfrac{3\pi-14}{3} $$
Hence, this is the answer.
Consider the given curves.
$$y=xe$$
$$y=-xe$$
$$x=1$$
Therefore, area bounded by these curves
$$ =\int_{0}^{1}{\left( xe-\left( -xe \right) \right)}dx $$
$$ =2e\int_{0}^{1}{x}dx $$
$$ =2e\left[ \dfrac{{{x}^{2}}}{2} \right]_{0}^{1} $$
$$ =e\left[ {{x}^{2}} \right]_{0}^{1} $$
$$ =e\left[ {{1}^{2}}-0 \right] $$
$$ =e $$
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