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

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

$$Curveisy=e^{-x^2}.$$$$Ans:a,b,d.$$

Answered by tanmay last updated on 28/Apr/19

$${\int }_0^1{e}^{-x^2}dx$$$${\int }_0^1(1-x^2+\frac{x^4}{2!}-\frac{x^6}{3!}+\frac{x^8}{4!}-...)dx$$$$\mid x-\frac{x^3}{3}+\frac{x^5}{5\times 2!}-\frac{x^7}{7\times 3!}+\frac{x^9}{9\times 4!}-...{\mid }_0^1$$$$=(1-\frac{1}{3}+\frac{1}{5\times 2!}-\frac{1}{7\times 3!}+\frac{1}{9\times 4!}....)$$$$=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{n-1}}{(2n-1)(n-1)!}$$$$waitiamtryingtofindthevalueof\mathrm{\Sigma }$$

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

$$thankusir.$$$$plcheckmycommentinQ.58648.$$

Commented by tanmay last updated on 28/Apr/19

$$foundtheanswerinbooksosharing...$$

Commented by tanmay last updated on 28/Apr/19

Commented by tanmay last updated on 28/Apr/19

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