For U_n = intlimits_0^1 x^n (2 - x)^n , dx an | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given $U_n = \int\limits_0^1 x^n (2 - x)^n \, dx$ and $V_n = \int\limits_0^1 x^n (1 - x)^n \, dx$. In the integral for $U_n$,substitute $x = 2t$,s

Vedclass · Accepted Answer

$U_n = 2^{2n} V_n$

Vedclass · Answer

(C) Given $U_n = \int\limits_0^1 x^n (2 - x)^n \, dx$ and $V_n = \int\limits_0^1 x^n (1 - x)^n \, dx$. In the integral for $U_n$,substitute $x = 2t$,so $dx = 2 \, dt$. When $x = 0, t = 0$ and when $x = 1, t = 1/2$. $U_n = \int\limits_0^{1/2} (2t)^n (2 - 2t)^n (2 \, dt) = \int\limits_0^{1/2} 2^n t^n 2^n (1 - t)^n (2 \, dt) = 2^{2n+1} \int\limits_0^{1/2} t^n (1 - t)^n \, dt$. Now,consider $V_n = \int\limits_0^1 x^n (1 - x)^n \, dx$. Using the property $\int\limits_0^{2a} f(x) \, dx = 2 \int\limits_0^a f(x) \, dx$ if $f(2a - x) = f(x)$,we check $f(x) = x^n (1 - x)^n$. Here $f(1 - x) = (1 - x)^n (1 - (1 - x))^n = (1 - x)^n x^n = f(x)$. Thus,$V_n = 2 \int\limits_0^{1/2} x^n (1 - x)^n \, dx$. Comparing $U_n$ and $V_n$,we see that $U_n = 2^{2n} \left( 2 \int\limits_0^{1/2} t^n (1 - t)^n \, dt \right) = 2^{2n} V_n$.

For $U_n = \int\limits_0^1 x^n (2 - x)^n \, dx$ and $V_n = \int\limits_0^1 x^n (1 - x)^n \, dx$,where $n \in N$,which of the following statements is true?

Similar Questions

$\int_{-1/2}^{1/2} \log \left(\frac{1+x}{1-x}\right) dx=$

The value of the definite integral $\int_{19}^{37} (\{x\}^2 + 3 \sin(2\pi x)) \, dx$,where $\{x\}$ denotes the fractional part function.

The value of the integral $\int_0^{\frac{\pi}{2}} \frac{\sqrt{\cot x}}{\sqrt{\cot x}+\sqrt{\tan x}} \,dx$ is

$\int\limits_a^b [x] \,dx + \int\limits_a^b [-x] \,dx$,where $[.]$ denotes the greatest integer function,is equal to:

The value of $\int_a^{a + (\pi /2)} (\sin^4 x + \cos^4 x) \, dx$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module