If the area of the region bounded by the curves y=x2, y=1x and the lines y = 0 and x = t (t>1) is 1 square unit, then t is equal to

y=x2,y=1x area =∫01 x2dx+∫1t 1xdx=x3301+[ln⁡(x)]1t=13+ln⁡(t) given that area enclosed is 1 ⇒13+ln⁡(t)=1 ⇒ln⁡(t)=23⇒t=e23

If the area of the region bounded by the curves y=x2, y=1x and the lines y = 0 and x = t (t>1) is 1 square unit, then t is equal to

y=x2,y=1x area =∫01 x2dx+∫1t 1xdx=x3301+[ln⁡(x)]1t=13+ln⁡(t) given that area enclosed is 1 ⇒13+ln⁡(t)=1 ⇒ln⁡(t)=23⇒t=e23

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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 square unit, then t is equal to

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e^3/2

$\frac{4}{3}$

$\frac{3}{2}$

e^2/3

answer is D.

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Detailed Solution

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

$area = \int_{0}^{1} x^{2} d x + \int_{1}^{t} \frac{1}{x} d x$

$= {[\frac{x^{3}}{3}]}_{0}^{1} + [\ln (x)]_{1}^{t}$

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

$given that area enclosed is 1 \begin{array}{l} \begin{aligned} \Rightarrow \frac{1}{3} + \ln (t) = 1 \\ \Rightarrow \ln (t) = \frac{2}{3} \end{aligned} \\ \Rightarrow t = e^{\frac{2}{3}} \end{array}$

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