Point of intersection of y = x² and y = $${1 \over x}$$.

put y = $${1 \over x}$$ in y = x², then we get,

$${1 \over x} = {x^2}$$

$$ \Rightarrow $$ $$\,\,\,$$ x³ $$-$$ 1 = 0

$$ \Rightarrow $$ $$\,\,\,$$ x = 1

$$\therefore\,\,\,$$ y = 1

$$\therefore\,\,\,$$ point B = (1, 1)

Area of region ABCDA

= $$\int\limits_0^1 {{x^2}} $$ dx + $$\int\limits_1^t {{1 \over x}} $$ dx

$$=$$ $$\left[ {{{{x^3}} \over 3}} \right]_0^1$$ + $$\left[ {\ell n\,x} \right]_1^t$$

= $${1 \over 3}$$ + $$\ell n\,t$$ $$-$$ $$\ell n$$ 1

= $${1 \over 3}$$ + $$\ell n\,t$$     [ as    $$\ell n$$ 1 = 0]

given this Area = 1 sq unit.

$$\therefore\,\,\,$$ $${1 \over 3}$$ + $$\ell n\,t$$ = 1

$$ \Rightarrow $$    $$\ell n\,t$$ = $${2 \over 3}$$

$$ \Rightarrow $$    t = e$${^{{2 \over 3}}}$$