calculate-0-1-e-x-1-x-dx-

Question Number 47182 by maxmathsup by imad last updated on 05/Nov/18

c a l c u l a t e \int_{0}^{1} e^{- x} \sqrt{1 - \sqrt{x}} d x

Commented by tanmay.chaudhury50@gmail.com last updated on 07/Nov/18

Commented by tanmay.chaudhury50@gmail.com last updated on 07/Nov/18

g r a p h o f e^{- x} \sqrt{1 - \sqrt{x}}

0.50 > \int_{0}^{1} e^{- x} \sqrt{1 - \sqrt{x}} d x > 0

Commented by maxmathsup by imad last updated on 11/Nov/18

l e t d e t e r m i n e a a p p r o x i m a t v a l u e o f A = \int_{0}^{1} e^{- x} \sqrt{1 - \sqrt{x}} d x w e h a v e

e^{u} = 1 + \frac{u}{1!} + \frac{u^{2}}{2!} + \dots . \Rightarrow e^{- x} = 1 - x + \frac{x^{2}}{2} - \dots \Rightarrow 0 ⩽ e^{- x} ⩽ 1 - x + \frac{x^{2}}{2} \Rightarrow

0 ⩽ e^{- x} \sqrt{1 - \sqrt{x}} ⩽ (1 - x + \frac{x^{2}}{2}) \sqrt{1 - \sqrt{x}} \Rightarrow 0 ⩽ \int_{0}^{1} e^{- x} \sqrt{1 - \sqrt{x}} d x ⩽ \int_{0}^{1} (1 - x + \frac{x^{2}}{2}) \sqrt{1 - \sqrt{x}} d x

c h a n g e m e n t \sqrt{1 - \sqrt{x}} = t g i v e 1 - \sqrt{x} = t^{2} \Rightarrow \sqrt{x} = 1 - t^{2} \Rightarrow x = {(1 - t^{2})}^{2} = t^{4} - 2 t^{2} + 1 \Rightarrow

\int_{0}^{1} (1 - x + \frac{x^{2}}{2}) \sqrt{1 - \sqrt{x}} d x = - \int_{0}^{1} (1 - t^{4} + 2 t^{2} - 1 + \frac{{(t^{4} - 2 t^{2} + 1)}^{2}}{2}) t (4 t^{3} - 4 t) d t

= - 2 \int_{0}^{1} (- 2 t^{4} + 4 t^{2} + (t^{8} + 4 t^{4} + 1 + 2 (- 2 t^{6} + t^{4} - 2 t^{2}) (t^{4} - t^{2}) d t

= - 2 \int_{0}^{1} (- 2 t^{4} + 4 t^{2} + t^{8} + 6 t^{4} - 4 t^{6} - 4 t^{2} + 1) (t^{4} - t^{2}) d t

= - 2 \int_{0}^{1} (t^{8} - 4 t^{6} - 2 t^{4} - 4 t^{2} + 1) (t^{4} - t^{2}) d t

= - 2 \int_{0}^{1} (t^{12} - t^{10} - 4 t^{10} + 4 t^{8} - 2 t^{8} + 2 t^{6} - 4 t^{6} + 4 t^{4}) d t

= - 2 \int_{0}^{1} (t^{12} - 5 t^{10} + 2 t^{8} - 2 t^{6} + 4 t^{4}) d t

= - 2 {[\frac{t^{13}}{13} - \frac{5 t^{11}}{11} + \frac{2 t^{9}}{9} - \frac{2 t^{7}}{7} + \frac{4 t^{5}}{5}]}_{0}^{1}

= - 2 (\frac{1}{13} - \frac{5}{11} + \frac{2}{9} - \frac{2}{7} + \frac{4}{5}) = - \frac{2}{13} + \frac{10}{11} - \frac{4}{9} + \frac{4}{7} - \frac{8}{5} = α_{0} \Rightarrow 0 < A ⩽ α_{0}