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JEE Main 201816 AprilOnline
Exam
JEE Main
Year
2018
Date
16 April
Mode
Online
Subject
Mathematics
Chapter
Area Under Curves

JEE Main 2018 · 16 April · Online · Mathematics · Area Under Curves

If the area of the region bounded by the curves, y = x ^ 2 , y =...

If the area of the region bounded by the curves, y=x2,y=1xandthelinesy = {x}^{2}, y = \dfrac{1}{x} and the linesy = 0andandx = t(t>1)\left(t > 1\right) is 1sq.unit1sq . unit, then tt is equal toIf the area of the region bounded by the curves, y = x ^ 2 , y = (1)/(x) and the lines y = 0 and x = t (t > 1) is 1sq .

Options

  1. Ae23{e}^{\dfrac{2}{3}}
  2. Be32{e}^{\dfrac{3}{2}}
  3. C32\dfrac{3}{2}
  4. D43\dfrac{4}{3}

Correct answer

A. e23{e}^{\dfrac{2}{3}}

Step-by-step solution

Area =∫01x2dx+∫1t1xdx=1= \mathop{∫}_{0}^{1}{x}^{2}dx + \mathop{∫}_{1}^{t}\dfrac{1}{x}dx = 1

If the area of the region bounded by the curves, y = x ^ 2 , y = (1)/(x) and the lines y = 0 and x = t (t > 1) is 1sq .

⇒13+ln⁡(t)=1⇒t=e23⇒ \dfrac{1}{3} + \ln \left(t\right) = 1 ⇒ t = e\dfrac{2}{3}

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