Question

Area bounded by $y=x^3,y=8$ and $x=0$ is ____

• A. $2$ sq. units
• B. $14$ sq. units
• C. $12$ sq. units
• D. $6$ sq. units

Answer 1

The correct option is C $12$ sq. units

Let,

$f\left(x\right)=x^3$ and $g\left(x\right)=8$

Given that,

$x=0$

In order to find second interval, we have to find the point of intersection of $y=x^3$ and $y=8$

Hence, $8=x^3$ $\therefore x=2$

hence the interval is, $[0,2]$

Area = ${\int }_a^b\left[f\right(x)-g(x\left)\right]dx$ where, $f\left(x\right)>g\left(x\right)$

In the given example, $g\left(x\right)>f\left(x\right)$

$\therefore$ Area = ${\int }_0^2[8-x^3]dx$

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

$\therefore A=[16-4]-[0-0]$

$\therefore A=12$ sq.units