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
$$\int_{\mathrm{0}} ^{\mathrm{1}} {e}^{−{x}^{\mathrm{2}} } {dx} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{x}^{\mathrm{2}} +\frac{{x}^{\mathrm{4}} }{\mathrm{2}!}−\frac{{x}^{\mathrm{6}} }{\mathrm{3}!}+\frac{{x}^{\mathrm{8}} }{\mathrm{4}!}−...\right){dx} \\ $$$$\mid{x}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}}+\frac{{x}^{\mathrm{5}} }{\mathrm{5}×\mathrm{2}!}−\frac{{x}^{\mathrm{7}} }{\mathrm{7}×\mathrm{3}!}+\frac{{x}^{\mathrm{9}} }{\mathrm{9}×\mathrm{4}!}−...\mid_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{5}×\mathrm{2}!}−\frac{\mathrm{1}}{\mathrm{7}×\mathrm{3}!}+\frac{\mathrm{1}}{\mathrm{9}×\mathrm{4}!}....\right) \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{\left(\mathrm{2}{n}−\mathrm{1}\right)\left({n}−\mathrm{1}\right)!} \\ $$$${wait}\:\:{i}\:{am}\:{trying}\:{to}\:{find}\:{the}\:{value}\:{of}\:\Sigma \\ $$
Commented by rahul 19 last updated on 28/Apr/19
$${thank}\:{u}\:{sir}. \\ $$$${pl}\:{check}\:{my}\:{comment}\:{in}\:{Q}.\mathrm{58648}. \\ $$
Commented by tanmay last updated on 28/Apr/19
$${found}\:{the}\:{answer}\:{in}\:{book}\:{so}\:{sharing}... \\ $$
Commented by tanmay last updated on 28/Apr/19
Commented by tanmay last updated on 28/Apr/19
