Question

Area bounded by tangent to curve $y=\underset{x^2}{\overset{x^3}{\int }}\frac{1}{\sqrt{1+t^2}}dt$ 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 $=(1,0)$
$\Theta y=\underset{x^2}{\overset{x^3}{\int }}\frac{1}{\sqrt{1+t^2}}dt$
$\Rightarrow \frac{dy}{dx}=\frac{1}{\sqrt{1+x^6}}\cdot 3x^2-\frac{1}{\sqrt{1+x^4}}\cdot 2x$
$\Rightarrow {\left(\frac{dy}{dx}\right)}_{(1,0)}=\frac{1}{\sqrt{2}}$
$\therefore$ Equation of tangent will be

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