Question

The area enclosed by $y=x^3$, its normal at $(1,1)$ and $x$ axis is equal to

• A. $\frac{7}{4}$ sq. units
• B. $\frac{9}{4}$ sq. units
• C. $\frac{5}{4}$sq. units
• D. $\frac{3}{4}$ sq. units

Answer 1

The correct option is A $\frac{7}{4}$ sq. units

$y=x^3,\frac{dy}{dx}=3x^2{\left(\frac{dy}{dx}\right)}_{(1,1)}=3$
Normal at $P(1,1)$ is $y-1=-\frac{1}{3}(x-1)$
$3y+x=4\cdots \left(i\right)$
So intersecting point of normal at $x$ -axis is $(4,0)$
Required Area
$=\underset{0}{\overset{1}{\int }}{x}^{3}dx+\frac{1}{2}(3\times 1)={\left[\frac{x^4}{4}\right]}_0^1+\frac{3}{2}$
$=\frac{7}{4}$sq. units