Question

If the area of the region bounded by the curves, $y=x^2,y=\frac{1}{x}$ and the lines $y=0$ and $x=t(t>1)$ is $1$ sq. unit, then t is equal to?

• A. $\frac{4}{3}$
• B. $e^{2/3}$
• C. $\frac{3}{2}$
• D. $e^{3/2}$

Answer 1

Answer: (B)

$e^{2/3}$

Given, $y=x^2,y=\frac{1}{x}$

Area bounded by the curves is the region ABCD.

Therefore, area $={\int }_0^1{x}^{2}dx+{\int }_1^t\frac{1}{x}dx$

$={\left[\frac{x^3}{3}\right]}_0^1+{\left[\ln \right(x\left)\right]}_1^t$

$=\frac{1}{3}+\ln \left(t\right)$

It is given that area enclosed is $1$

$\Rightarrow \frac{1}{3}+\ln \left(t\right)=1$

$\Rightarrow \ln \left(t\right)=\frac{2}{3}$

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

Hence, answer is option (B)