Mathematics - Area Under Curves Question with Solution | TestHub

The required area is bounded by the curve $y=x^3$, the line $y=8$, and the y-axis ($x=0$).

The point of intersection of $y=x^3$ and $y=8$ in the first quadrant is $x=2$.

The area can be calculated by integrating with respect to $x$ from $x=0$ to $x=2$:

$A={\int }_0^2(8-x^3)dx$

$A={\left[8x-\frac{x^4}{4}\right]}_0^2$

$A=16-\frac{16}{4}$

$A=16-4=12$

Alternatively, integrating with respect to $y$ from $y=0$ to $y=8$:

$A={\int }_0^{8}xdy={\int }_0^8{y}^{1/3}dy$

$A={\left[\frac{3}{4}{y}^{4/3}\right]}_0^8$

$A=\frac{3}{4}(8^{4/3})=\frac{3}{4}(16)=12$

Answer: $12$