Question-58717

Question Number 58717 by rahul 19 last updated on 28/Apr/19

Commented by rahul 19 last updated on 28/Apr/19

C u r v e i s y = e^{- x^{2}} .

A n s : a, b, d .

Answered by tanmay last updated on 28/Apr/19

\int_{0}^{1} e^{- x^{2}} d x

\int_{0}^{1} (1 - x^{2} + \frac{x^{4}}{2!} - \frac{x^{6}}{3!} + \frac{x^{8}}{4!} - \dots) d x

∣ x - \frac{x^{3}}{3} + \frac{x^{5}}{5 \times 2!} - \frac{x^{7}}{7 \times 3!} + \frac{x^{9}}{9 \times 4!} - \dots ∣_{0}^{1}

= (1 - \frac{1}{3} + \frac{1}{5 \times 2!} - \frac{1}{7 \times 3!} + \frac{1}{9 \times 4!} \dots .)

= \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n - 1}}{(2 n - 1) (n - 1)!}

w a i t i a m t r y i n g t o f i n d t h e v a l u e o f Σ

Commented by rahul 19 last updated on 28/Apr/19

t h a n k u s i r .

p l c h e c k m y c o m m e n t i n Q . 58648 .

Commented by tanmay last updated on 28/Apr/19

f o u n d t h e a n s w e r i n b o o k s o s h a r i n g \dots

Commented by tanmay last updated on 28/Apr/19

Commented by tanmay last updated on 28/Apr/19

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