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

Answer 1

The correct option is D $e^{\frac{2}{3}}$

$Area={\int }_0^1{x}^{2}dx+{\int }_1^t\frac{1}{x}dx$
$Area={\left(\frac{x^3}{3}\right)}_0^1+{\left(\ln x\right)}_1^t$
$1=(\frac{1}{3}-\frac{0}{3})+(\ln t-0)$
$1-\frac{1}{3}=\ln t$
$\frac{2}{3}=\ln t$
$t=e^{\frac{2}{3}}$