Home » Blog » Prove an identity of integral equations Prove an identity of integral equations by RoRi October 19, 2015 For positive integers prove Proof. Let Then, Related
June 8, 2022 at 1:49 pm José Muñoz says: \[ u = 1-x \qquad \implies \qquad du = -dx. \] Then, \begin{align*} – \int_{u(0)}^{u(1)} (1-u)^m u^n \, du \\ &= -\int_1^0 (1-u)^n u^m \, du \int_{u(0)}^{u(1)} (1-u)^m u^n \, du \\ &= \int_1^0 (1-u)^n u^m \, du We can change back to x \int_0^1 x^m (1-x)^n \, dx.\\ &= \int_0^1 x^n (1-x)^m \, dx. \qquad \blacksquare \end{align*}
prove![\[ \int_0^1 x^m (1-x)^n \, dx = \int_0^1 x^n (1-x)^m \, dx. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-a89d899cd325d30d40fa792d1a85397d_l3.png "Rendered by QuickLaTeX.com")
![\[ u = 1-x \qquad \implies \qquad du = -dx. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-d750e2e50730611ddc7c3bbcdc17de2b_l3.png "Rendered by QuickLaTeX.com")
prove
\[ u = 1-x \qquad \implies \qquad du = -dx. \]
Then,
\begin{align*}
– \int_{u(0)}^{u(1)} (1-u)^m u^n \, du \\ &= -\int_1^0 (1-u)^n u^m \, du
\int_{u(0)}^{u(1)} (1-u)^m u^n \, du \\ &= \int_1^0 (1-u)^n u^m \, du
We can change back to x
\int_0^1 x^m (1-x)^n \, dx.\\ &= \int_0^1 x^n (1-x)^m \, dx. \qquad \blacksquare \end{align*}