By using the properties of definite integrals,evaluate the integral $\int_{0}^{1} x(1-x)^{n} d x$.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
Let $I = \int_{0}^{1} x(1-x)^{n} d x$.
Using the property $\int_{0}^{a} f(x) d x = \int_{0}^{a} f(a-x) d x$,we get:
$I = \int_{0}^{1} (1-x)(1-(1-x))^{n} d x$
$I = \int_{0}^{1} (1-x)(x)^{n} d x$
$I = \int_{0}^{1} (x^{n} - x^{n+1}) d x$
Integrating term by term:
$I = \left[ \frac{x^{n+1}}{n+1} - \frac{x^{n+2}}{n+2} \right]_{0}^{1}$
Evaluating at the limits:
$I = \left( \frac{1^{n+1}}{n+1} - \frac{1^{n+2}}{n+2} \right) - (0 - 0)$
$I = \frac{1}{n+1} - \frac{1}{n+2}$
$I = \frac{(n+2) - (n+1)}{(n+1)(n+2)}$
$I = \frac{1}{(n+1)(n+2)}$

Explore More

Similar Questions

Let $g(x) = \int_{0}^{x} f(t) dt$,where $f$ is a continuous function in $[0, 3]$ such that $\frac{1}{3} \leq f(t) \leq 1$ for all $t \in [0, 1]$ and $0 \leq f(t) \leq \frac{1}{2}$ for all $t \in (1, 3]$. The largest possible interval in which $g(3)$ lies is:

The value of the integral $\int_{0}^{1} x \cot^{-1}(1 - x^2 + x^4) dx$ is

Let $f:(0,2) \rightarrow R$ be defined as $f(x) = \log_{2}\left(1+\tan\left(\frac{\pi x}{4}\right)\right)$. Then,$\lim_{n \rightarrow \infty} \frac{2}{n}\left(f\left(\frac{1}{n}\right)+f\left(\frac{2}{n}\right)+\ldots+f(1)\right)$ is equal to

Value of the definite integral $\int_{-3}^{1} (2(t+1)^5 - 5(t+1)^3 + t + 3) dt$ is equal to

$\int_{-\pi / 2}^{\pi / 2} \sin |x| \, dx$ is equal to

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo