Question

Find the total area between the curve $y=x^3$ and the x-axis between $x=-2$ and $x=2$.

• A. $2$
• B. $4$
• C. $6$
• D. $8$

Answer 1

The correct option is D $8$

Required Area = ${\int }_0^2{x}^{3}dx+{\int }_{-2}^0{x}^{3}dx$

$={\left[\frac{x^4}{4}\right]}_0^2+{\left[\frac{x^4}{4}\right]}_{-2}^0$

$=\frac{16}{4}+(-\frac{16}{4})$

We would take $-\frac{16}{4}$ as positive as it is area.

Thus, required area = $4+4=8sq.units$