prove-that-a-a-dx-x-n-1-x-2n-1-a-

Question Number 207620 by Ghisom last updated on 21/May/24

prove that \underset{- a}{\int^{a}} \frac{d x}{x^{n} + 1 + \sqrt{x^{2 n} + 1}} = a

Answered by Berbere last updated on 21/May/24

\int_{- a}^{a} \frac{x^{n} + 1 - \sqrt{1 + x^{2 n}}}{2 x^{n}} = U_{n}

= \int_{- a}^{a} \frac{1}{2} + \frac{1 - \sqrt{1 + x^{2 n}}}{2 x^{n}} = a + \int_{- a}^{a} \frac{1}{2} + \int_{- a}^{a} \frac{1 - \sqrt{1 + x^{2 n}}}{2 x^{n}} d x

= a + \frac{1}{2} \int_{- a}^{a} \frac{1 - \sqrt{1 + x^{2 n}}}{2 x^{n}} d x

\Rightarrow \forall n \in N \int_{- a}^{a} \frac{1 - \sqrt{1 + x^{2 n}}}{2 x^{n}} = 0 y o u c a n s e e i t i f n = 2 k

\frac{1 - \sqrt{1 + x^{2 n}}}{2 x^{n}} 0 <; \forall x \in R - {0}

\Rightarrow \int_{- a}^{a} \frac{1 - \sqrt{1 + x^{2 n}}}{2 x^{n}} < 0 \neq 0 o n l y i f a = 0

T h e r e s u l t a i s t r u e i f n = 2 k + 1; k \in N

\overset{*}{f} = {\begin{cases} \frac{1 - \sqrt{1 + x^{2 n}}}{2 x^{n}} = x \neq 0; n = 2 k + 1; k \in N \\ = 0 i f x = 0 \end{cases}

\overset{*}{f} (- x) = - f (x); \forall x \in R

\int_{- a}^{a} \overset{*}{f} (x) d x = 0 \Rightarrow U_{2 n + 1} = a

Commented by Ghisom last updated on 21/May/24

thank you

Facebook Tweet Pin