Question-187890

Question Number 187890 by Rupesh123 last updated on 23/Feb/23

Answered by aleks041103 last updated on 23/Feb/23

I = \underset{- 1}{\int^{1}} \frac{4 \sqrt{1 - x^{2}}}{1 + 2^{x}} d x

u = - x, d u = - d x

I = \int_{1}^{- 1} \frac{4 \sqrt{1 - {(- u)}^{2}}}{1 + 2^{- u}} (- d u) =

= \int_{- 1}^{1} \frac{2^{u}}{1 + 2^{u}} 4 \sqrt{1 - u^{2}} d u =

= \int_{- 1}^{1} 4 \sqrt{1 - u^{2}} d u - \int_{- 1}^{1} \frac{4 \sqrt{1 - u^{2}}}{1 + 2^{u}} d u

\Rightarrow I = 2 \underset{- 1}{\int^{1}} \sqrt{1 - x^{2}} d x = π

\Rightarrow I = π

Commented by Rupesh123 last updated on 23/Feb/23

Perfect ��