Mathematics - Area Under Curves Question with Solution | TestHub

The region is bounded by the curves $y=x^2$, $xy=8$, and $y=1$ in the first quadrant (as the area must be finite).

To find the intersection points of the curves:
Intersection of $y=x^2$ and $xy=8$:
$x(x^2)=8\Rightarrow x^3=8\Rightarrow x=2$
At $x=2$, $y=4$.

Intersection of $y=x^2$ and $y=1$:
$x^2=1\Rightarrow x=1$ (taking $x>0$).

Intersection of $xy=8$ and $y=1$:
$x(1)=8\Rightarrow x=8$.

The area can be calculated by integrating with respect to $y$ from $y=1$ to $y=4$. For a given $y$, the value of $x$ ranges from the parabola $x=\sqrt{y}$ to the hyperbola $x=\frac{8}{y}$.

Required Area $={\int }_1^4\left(\frac{8}{y}-\sqrt{y}\right)dy$

$={\left[8\log y-\frac{2}{3}{y}^{3/2}\right]}_1^4$

$=\left(8\log 4-\frac{2}{3}(4{)}^{3/2}\right)-\left(8\log 1-\frac{2}{3}(1{)}^{3/2}\right)$

$=\left(16\log 2-\frac{2}{3}(8)\right)-\left(0-\frac{2}{3}\right)$

$=16\log 2-\frac{16}{3}+\frac{2}{3}$

$=16\log 2-\frac{14}{3}$

Answer: $16\log 2-\frac{14}{3}$