Question-58717

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

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

$${Curve}\:{is}\:{y}={e}^{−{x}^{\mathrm{2}} } . \\ $$$${Ans}:{a},{b},{d}. \\ $$

Answered by tanmay last updated on 28/Apr/19

∫_0 ^1 e^(−x^2 ) dx ∫_0 ^1 (1−x^2 +(x^4 /(2!))−(x^6 /(3!))+(x^8 /(4!))−...)dx ∣x−(x^3 /3)+(x^5 /(5×2!))−(x^7 /(7×3!))+(x^9 /(9×4!))−...∣_0 ^1 =(1−(1/3)+(1/(5×2!))−(1/(7×3!))+(1/(9×4!))....) =Σ_(n=1) ^∞ (((−1)^(n−1) )/((2n−1)(n−1)!)) wait i am trying to find the value of Σ

$$\int_{\mathrm{0}} ^{\mathrm{1}} {e}^{−{x}^{\mathrm{2}} } {dx} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{x}^{\mathrm{2}} +\frac{{x}^{\mathrm{4}} }{\mathrm{2}!}−\frac{{x}^{\mathrm{6}} }{\mathrm{3}!}+\frac{{x}^{\mathrm{8}} }{\mathrm{4}!}−…\right){dx} \\ $$$$\mid{x}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}}+\frac{{x}^{\mathrm{5}} }{\mathrm{5}×\mathrm{2}!}−\frac{{x}^{\mathrm{7}} }{\mathrm{7}×\mathrm{3}!}+\frac{{x}^{\mathrm{9}} }{\mathrm{9}×\mathrm{4}!}−…\mid_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{5}×\mathrm{2}!}−\frac{\mathrm{1}}{\mathrm{7}×\mathrm{3}!}+\frac{\mathrm{1}}{\mathrm{9}×\mathrm{4}!}….\right) \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{\left(\mathrm{2}{n}−\mathrm{1}\right)\left({n}−\mathrm{1}\right)!} \\ $$$${wait}\:\:{i}\:{am}\:{trying}\:{to}\:{find}\:{the}\:{value}\:{of}\:\Sigma \\ $$

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