$\int_{0}^{1} x {[1 - x]}^{11} d x = . . . . . . . .$

- \frac{1}{132}

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

- \frac{1}{156}

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

- \frac{1}{121}

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

- \frac{1}{12}

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

Open in App

Solution

The correct option is A $- \frac{1}{156}$
$\to I = \int_{0}^{1} x {(1 - x)}^{11} d x$
Take $1 - x$ is place of $x$
$= \int_{0}^{1} (1 - x) {[1 - (1 - x)]}^{11} d x$
$= - \int_{0}^{1} (1 - x) x^{11} d x = - \int_{0}^{1} (x^{11} - x^{12}) d x = - {[\frac{x^{12}}{12} - \frac{x^{13}}{13}]}_{0}^{1} = \frac{1}{13} - \frac{1}{12}$
$∴ I = - \frac{1}{156}$

Suggest Corrections

Similar questions

A = \{x \in R : x \geq 1\}

f : A \to A

f (x) = 2^{x (x - 1)}, is

{(\frac{1}{2})}^{x (x - 1)}

\frac{1}{2} \{1 + \sqrt{1 + 4 \log_{2} x}\}

\frac{1}{2} \{1 - \sqrt{1 + 4 \log_{2} x}\}

\int_{0}^{1} \tan^{- 1} (\frac{2 x}{1 - x^{2}}) d x

\lim_{x \to 0} \frac{\sqrt{1 + x} - 1}{x}

\frac{1}{2}

L e t I = 1 \int 0 \sqrt{\frac{1 + \sqrt{x}}{1 - \sqrt{x}}} d x

J = 1 \int 0 \sqrt{\frac{1 - \sqrt{x}}{1 + \sqrt{x}}} d x,

{lim}_{x \to 0} \frac{\sqrt{1 + x} - 1}{x} = {lim}_{x \to 0} \frac{(1 + x) \frac{1}{n} - 1}{(x + 1) - 1} =

Join BYJU'S Learning Program