Find-area-between-by-y-1-and-y-1-x-2-1-x-2-

Question Number 29202 by ajfour last updated on 05/Feb/18

F i n d a r e a b e t w e e n b y y = 1 a n d

y = \frac{1 - x^{2}}{1 + x^{2}} .

Answered by mrW2 last updated on 05/Feb/18

y = \frac{1 - x^{2}}{1 + x^{2}} = \frac{2}{1 + x^{2}} - 1

i f x \to \pm \infty, y \to y_{m i n} = - 1

i f x = 0, y = y_{m a x} = 1

i f x = \pm 1, y = 0

A r e a b e t w e e n y = - 1 a n d y = \frac{1 - x^{2}}{1 + x^{2}} :

A = 2 \int_{0}^{+ \infty} (y + 1) d x

= 2 \int_{0}^{+ \infty} (\frac{2}{1 + x^{2}}) d x

= 4 \int_{0}^{+ \infty} (\frac{1}{1 + x^{2}}) d x

= 4 {[\tan^{- 1} x]}_{0}^{+ \infty}

= 4 (\frac{π}{2} - 0)

= 2 π

Commented by mrW2 last updated on 05/Feb/18

A r e a b e t w e e n y = 1 a n d y = \frac{1 - x^{2}}{1 + x^{2}} i s \infty .

Commented by ajfour last updated on 05/Feb/18

I h a d i n t e n d e d t o a s k

\int_{- \infty}^{\infty} (\frac{x^{2} - 1}{x^{2} + 1}) d x S i r, t h i s i s a g a i n

= 2 π .

Commented by mrW2 last updated on 05/Feb/18

I s e e .

Facebook Tweet Pin