Question

Choose the correct answer
Area bounded by the curve $y=x^3$, the X-axis the coordinates x = - 2 and x = 1 si
(a) -9 (b) $-\frac{15}{4}$
(c) $\frac{15}{4}$ (d) $\frac{17}{4}$

Answer 1

Given curve is $y=x^3$
On putting x = - and y = - y, we get $y=x^3$
Therefore, the curve is symmetrical in opposite quadrant and passes through (0, 0).
$\therefore$ Required area
$={\int }_{-2}^1|x^3|dx={\int }_{-2}^0|x^3|dx+{\int }_0^1|x^3|dx$
$={\int }_{-2}^0(-x^3)dx+{\int }_0^1{x}^{3}dx$
$[\because |x^3|=\left\{\begin{array}{cc}{x}^3,& ifx\ge 0\\ -x^3,& ofx<0\end{array}\right\}]=-[\frac{x^4}{4}{]}_{-2}^0+[\frac{x^4}{4}{]}_0^1=-[0-\frac{(-2{)}^4}{4}]+[\frac{1}{4}-0]=-(-4)+\left(\frac{1}{4}\right)=\frac{17}{4}squnit$
So, the correct option is (d).