Question

Area bounded by the curve y = x³, the x-axis and the ordinates x = −2 and x = 1 is
(a) −9

(b) $\frac{-15}{4}$

(c) $\frac{15}{4}$

(d) $\frac{17}{4}$

Answer 1

$\left(\mathrm{d}\right)\frac{17}{4}$



$$\begin{gathered} x=-2\mathrm{and}x=1\mathrm{intersect}\mathrm{the}\mathrm{curve}y=x^3\mathrm{at}\mathrm{A}\left(-2,-8\right)\mathrm{and}\mathrm{B}\left(1,1\right)\mathrm{respectively} \\ \mathrm{If}P\left(x,y_1\right)\mathrm{lies}\mathrm{on}OA\&Q\left(x,y_2\right)\mathrm{lies}\mathrm{on}\mathrm{curve}OB \\ \mathrm{Then},{\mathrm{y}}_1>0\Rightarrow \left|y_1\right|=y_1 \\ \&{\mathrm{y}}_2<0\Rightarrow \left|y_2\right|=-y_2 \\ \mathrm{Area}\mathrm{of}\mathrm{curve}\mathrm{bound}\mathrm{by}\mathrm{the}\mathrm{two}\mathrm{lines}=\mathrm{shaded}\mathrm{area}\left(\mathit{O}\mathit{A}\mathit{D}\mathit{O}\right)+\mathrm{Shaded}\mathrm{area}\left(\mathit{O}\mathit{C}\mathit{B}\mathit{O}\right) \\ ={\int }_{-2}^0\left|y_2\right|dx+{\int }_0^1\left|y_1\right|dx \\ ={\int }_{-2}^0-y_{2}dx+{\int }_0^1{y}_{1}dx \\ ={\int }_{-2}^0-\left(x^3\right)dx+{\int }_0^1{x}^{3}dx \\ ={\left[-\frac{x^4}{4}\right]}_{-2}^0+{\left[\frac{x^4}{4}\right]}_0^1 \\ =0-\left(-\frac{16}{4}\right)+\left(\frac{1}{4}-0\right) \\ =4+\frac{1}{4} \\ =\frac{17}{4}\mathrm{sq}\mathrm{units} \end{gathered}$$