The area bounded by the curves $y = x (1 - l n x)$ and positive x-axis between $x = e^{- 1}$ and $x = e$ is

(\frac{e^{2} - 4 e^{- 2}}{5})

No worries! We‘ve got your back. Try BYJU‘S free classes today!

(\frac{e^{2} - 5 e^{- 2}}{4})

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

(\frac{4 e^{2} - e^{- 2}}{5})

No worries! We‘ve got your back. Try BYJU‘S free classes today!

(\frac{5 e^{2} - e^{- 2}}{4})

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $(\frac{e^{2} - 5 e^{- 2}}{4})$
$y = x (1 - ln x)$
Area under the curve $A = \int_{1 / e}^{e} y d x$
$= \int_{1 / e}^{e} x (1 - ln x) d x$
Integrating by parts, we get
$A = {[\frac{x^{2}}{2} - (ln x \frac{x^{2}}{2} - \frac{x^{2}}{4})]}_{1 / e}^{e}$
$= (\frac{e^{2} - 5 e^{- 2}}{4})$

Suggest Corrections

Similar questions

x = 1, x = 2,

x (y - e^{x}) = sin x

2 x y = 2 sin x + x^{3}

Area bounded by the curves $y = e^{x}, y = l o g_{e} x$ and the lines x = 0, y = 0, y = 1 is

y = (l n x)^{2}

x = e^{2}

Area bounded by the curve $y = x e^{x^{2}}$ x - axis and the ordinates x = 0, x = a

y = x e^{x^{2}}

x

x = 0

x = a

Join BYJU'S Learning Program