The required area is clearly on both the positive side of y-axis(i .e)in the Ist quadrant and negative side that is the IIIrd quadrant.

This is shown in the fig.

The required area is the area bounded by the curve $y=x^3$,the x-axis and the ordinate x=-2 and x=1.

Hence we can take the limits as -2 to 1.

$A=\int_{-2}^1x^3dx$

But this curve lies on both the positive side and the negative side of x-axis,let us split the limits as -2 to 0 and 0 to 1

$A=\int_{-2}^0x^3 dx+\int_0^1x^3 dx$

on integrating we get,

$A=\begin{bmatrix}\frac{x^4}{4}\end{bmatrix}_{-2}^0+\begin{bmatrix}\frac{x^4}{4}\end{bmatrix}_0^1$

on applying limits we get,

$A=\begin{bmatrix}0+\frac{(-2)^4}{4}\end{bmatrix}+\frac{1^4}{4}$ [we take $\frac{-16}{4}$ as $\frac{16}{4}$ because area cannot be negative]

$\;\;\;=\frac{16}{4}+\frac{1}{4}=\frac{17}{4}$sq.units.

Hence the required area is $\frac{17}{4}$sq.units.

The correct answer is D