Question

The area of the region bounded by the curve $y=x^3$, its tangent at $(1,1)$ and $x$-axis, is

• A. $\frac{1}{12}$ sq unit
• B. $\frac{1}{6}$ sq unit
• C. $\frac{2}{17}$ sq unit
• D. $\frac{2}{15}$ sq unit

Answer 1

The correct option is A $\frac{1}{12}$ sq unit

We have, $y=x^3$ and $A(1,1)$
$\therefore \frac{dy}{dx}=3x^2$

On putting $x=1$ in Eq. (i), we get
$\frac{dy}{dx}=3(1{)}^2=3$

$\therefore$ Equation of tangent at $A(1,1)$ is
$y-1=3(x-1)\Rightarrow y=3x-2$

$\therefore$ Required area
$={\int }_0^1{x}^{3}dx-{\int }_{2/3}^1(3x-2)dx$

$={\left[\frac{x^4}{4}\right]}_0^1-{[\frac{3x^2}{2}-2x]}_{2/3}^2$

$=\frac{1}{4}[(\frac{3}{2}-2)-(\frac{2}{3}-\frac{4}{3})]$

$=\frac{1}{12}$ sq unit