If int_{0}^{1} x^{m} (1 - x)^{n} dx = k int_{ | vedclass

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

Question

(D) We are given the integral $I = \int_{0}^{1} x^{m} (1 - x)^{n} dx$. Using the property of definite integrals,$\int_{0}^{a} f(x) dx = \int_{0}^{a} f

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(D) We are given the integral $I = \int_{0}^{1} x^{m} (1 - x)^{n} dx$. Using the property of definite integrals,$\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a - x) dx$,we substitute $x$ with $(1 - x)$: $I = \int_{0}^{1} (1 - x)^{m} (1 - (1 - x))^{n} dx$ $I = \int_{0}^{1} (1 - x)^{m} (x)^{n} dx$ $I = \int_{0}^{1} x^{n} (1 - x)^{m} dx$. Comparing this with the given equation $\int_{0}^{1} x^{m} (1 - x)^{n} dx = k \int_{0}^{1} x^{n} (1 - x)^{m} dx$,we see that $I = k \cdot I$. Since the integral is non-zero,we conclude that $k = 1$.

If $\int_{0}^{1} x^{m} (1 - x)^{n} dx = k \int_{0}^{1} x^{n} (1 - x)^{m} dx$,then the value of $k$ equals

Similar Questions

$\int_0^{\pi /2} \frac{x \sin x \cos x}{\cos^4 x + \sin^4 x} \, dx = $

$\frac{3}{25} \int_0^{25 \pi} \sqrt{|\cos x - \cos^3 x|} \, dx =$

$\int_0^\pi x f(\sin x) dx = $

$\int_{-1}^{1} \sin^{11} x \, dx$ is equal to

$\int_0^{\frac{\pi}{2}} \frac{\cos x \, dx}{\sqrt{1+\cos x \sin x}} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module