Area bounded by the curve $y=x^3$, the x-axis and the ordinates $x=2$and $x=1$is(A) $-9$ (B) $\frac{-15}{4}$ (C) $\frac{15}{4}$ (D) $\frac{17}{4}$

Solution: From, the given details, we can draw the diagram. Please refer video for the diagram.
As per diagram, required area will be,
${\int }_{-2}^0{x}^{3}dx+{\int }_0^1{x}^{3}dx$
$={\left|\frac{x^4}{4}\right|}_{-2}^0+{\left|\frac{x^4}{4}\right|}_0^1$
$=\frac{16}{4}+\frac{1}{4}=\frac{17}{4}$, which is the required area.