JEE Advanced 2025 Paper 2 · Mathematics · Area Under Curves · medium

Let R be the region enclosed by the curves y=ln x and y=(ln x)^2 and the vertical lines x=1 and x=e^2. The area of R is ______.

Numerical answer

Let

R

be the region enclosed by the curves

y=\ln x

and

y=(\ln x)^2

and the vertical lines

x=1

and

x=e^2

. The area of

R

is ______. Give the answer correct to two decimal places.

Show answer and solution

Answer: 4.95

The curves meet where

\ln x=\ln^2x

, i.e.

x=1

and

x=e

. On

[1,e]

\ln x\ge\ln^2x

; on

[e,e^2]

the order reverses. Use

\int\ln x\,dx=x\ln x-x

and

\int\ln^2x\,dx=x(\ln^2x-2\ln x+2)

. On

[1,e]

\int\ln x\,dx=1

and

\int\ln^2x\,dx=e-2

, so the area is

1-(e-2)=3-e

. On

[e,e^2]

\int\ln^2x=2e^2-e

\int\ln x=e^2

, area

=e^2-e

. Total

=3-2e+e^2=(e-1)^2+2

. Numerically

3-5.4366+7.3891=4.9525\approx4.95

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