Question

Area bounded by tangent to curve $y =\int\limits_{ x ^2}^{ x ^3} \frac{1}{\sqrt{1+ t ^2}} dt$ at $x =1$, with coordinate axes is equal to

• A. $\frac{1}{2 \sqrt{2}}$
• B. $\frac{1}{\sqrt{2}}$
• C. $\sqrt{2}$
• D. $2 \sqrt{2}$

Answer 1

Answer: (A)

Point of contact $=(1,0)$
$\Theta y=\int\limits_{x^2}^{x^3} \frac{1}{\sqrt{1+t^2}} d t $
$\Rightarrow \frac{d y}{d x}=\frac{1}{\sqrt{1+x^6}} \cdot 3 x^2-\frac{1}{\sqrt{1+x^4}} \cdot 2 x $
$\Rightarrow \left(\frac{d y}{d x}\right)_{(1,0)}=\frac{1}{\sqrt{2}}$
$\therefore $ Equation of tangent will be

$ y-0=\frac{1}{\sqrt{2}}(x-1) $
$\therefore \text { Area bounded }=\frac{1}{2} \times \frac{1}{\sqrt{2}} \times 1=\frac{1}{2 \sqrt{2}}$