Mathematics - Area Under Curves Question with Solution | TestHub

To draw the inequality, let us draw the equation
$xy=8$and$y=1$and$y=x^2$
For point of intersection
(i)$xy=8$and$y=1\Rightarrow A\left(\mathrm{8,1}\right)$
(ii)$xy=8$and$y=x^2 \Rightarrow x^3=8 \Rightarrow x=2 \Rightarrow B(2, 4)$
(iii)$y=x^2$and$y=1\Rightarrow C\left(1, 1\right)$and$D\left( 1, 1\right)$
Now$xy\le 8\Rightarrow$region which contains origin
$y\ge 1\Rightarrow$region above line$y=1$
$y\le x^2\Rightarrow$region outside the parabola
Now required area
Method I:
Using x-axis:
$A=\underset{1}{\overset{2}{\int }}\left(x^2-1\right)\cdot dx+\underset{2}{\overset{8}{\int }}\left(\frac{8}{x}-1\right)\cdot dx$
$A={\left[\frac{x^3}{3}-x\right]}_1^2+{\left[8\mathrm{l}\mathrm{n}x-x\right]}_2^8$
$A=-\frac{14}{3}+16\mathrm{l}\mathrm{n}2$
Method II:
Using y-axis:
$A=\underset{1}{\overset{4}{\int }}\left(\frac{8}{y}-\sqrt{y}\right)\cdot dy$
$A={\left[8\mathrm{l}\mathrm{n}y-\frac{2}{3}{y}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right]}_1^4\Rightarrow A=-\frac{14}{3}+16\mathrm{l}\mathrm{n}2$
Note: The question should include bounded area term as in$2^{\mathrm{n}\mathrm{d}}$quadrant there exist a area which satisfy the inequality and is unbounded.