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0-1-1-e-x-2-dx-

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Question Number 53384 by gunawan last updated on 21/Jan/19
∫_( 0) ^1 (1+e^(−x^2 ) ) dx =
$$\:\underset{\:\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\left(\mathrm{1}+{e}^{−{x}^{\mathrm{2}} } \right)\:{dx}\:= \\ $$
Commented by Abdo msup. last updated on 21/Jan/19
we have e^u =Σ_(n=0) ^∞ (u^n /(n!)) ⇒e^(−x^2 ) = Σ_(n=0) ^∞ (((−1)^n x^(2n) )/(n!)) ⇒ ∫_0 ^1 (1+e^(−x^2 ) dx =1+∫_0 ^1 e^(−x^2 ) dx =1+Σ_(n=0) ^∞ (((−1)^n )/(n!)) ∫_0 ^1 x^(2n) dx =1+Σ_(n=0) ^∞ (((−1)^n )/((2n+1)n!)) =1+1 −(1/3) +(1/(5(2!))) −(1/(7(3!))) +(1/(9(4!))) −(1/(11(5!))) +... =(5/3) +(1/(10)) −(1/(42)) +.... we get best approximation for this integral if we take 10 terms...
$${we}\:{have}\:{e}^{{u}} =\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{u}^{{n}} }{{n}!}\:\Rightarrow{e}^{−{x}^{\mathrm{2}} } =\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} {x}^{\mathrm{2}{n}} }{{n}!}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}+{e}^{−{x}^{\mathrm{2}} } {dx}\:=\mathrm{1}+\int_{\mathrm{0}} ^{\mathrm{1}} \:{e}^{−{x}^{\mathrm{2}} } {dx}\right. \\ $$$$=\mathrm{1}+\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}^{\mathrm{2}{n}} {dx} \\ $$$$=\mathrm{1}+\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right){n}!} \\ $$$$=\mathrm{1}+\mathrm{1}\:−\frac{\mathrm{1}}{\mathrm{3}}\:+\frac{\mathrm{1}}{\mathrm{5}\left(\mathrm{2}!\right)}\:−\frac{\mathrm{1}}{\mathrm{7}\left(\mathrm{3}!\right)}\:+\frac{\mathrm{1}}{\mathrm{9}\left(\mathrm{4}!\right)}\:−\frac{\mathrm{1}}{\mathrm{11}\left(\mathrm{5}!\right)}\:+… \\ $$$$=\frac{\mathrm{5}}{\mathrm{3}}\:+\frac{\mathrm{1}}{\mathrm{10}}\:−\frac{\mathrm{1}}{\mathrm{42}}\:+…. \\ $$$${we}\:{get}\:{best}\:{approximation}\:{for}\:{this}\:{integral} \\ $$$${if}\:{we}\:{take}\:\mathrm{10}\:{terms}… \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 21/Jan/19
∫_0 ^1 (1+(1/e^x^2 ))dx this intrgal can be solved withthe help of graph f(x)=1+(1/e^x^2 ) f(0)=2 f(1)=1+e^(−1) =0<∫_0 ^1 (1+(1/e^x^2 ))<(1/2){f(0)+f(1)}×1 0<I<(1/2)(2+1+e^(−1) ) 0<I<(1/2)(3+(1/e)) 0<I<1.68 but from attached graph the area is A+B but we found A=(1/2)(3+(1/e)) so actual value of ∫_0 ^1 (1+(1/e^x^2 ))dx=A+B
$$\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}+\frac{\mathrm{1}}{{e}^{{x}^{\mathrm{2}} } }\right){dx} \\ $$$${this}\:{intrgal}\:{can}\:{be}\:{solved}\:{withthe}\:{help}\:{of}\:{graph} \\ $$$${f}\left({x}\right)=\mathrm{1}+\frac{\mathrm{1}}{{e}^{{x}^{\mathrm{2}} } } \\ $$$${f}\left(\mathrm{0}\right)=\mathrm{2} \\ $$$${f}\left(\mathrm{1}\right)=\mathrm{1}+{e}^{−\mathrm{1}} \\ $$$$=\mathrm{0}<\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}+\frac{\mathrm{1}}{{e}^{{x}^{\mathrm{2}} } }\right)<\frac{\mathrm{1}}{\mathrm{2}}\left\{{f}\left(\mathrm{0}\right)+{f}\left(\mathrm{1}\right)\right\}×\mathrm{1} \\ $$$$\:\:\mathrm{0}<{I}<\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}+\mathrm{1}+{e}^{−\mathrm{1}} \right) \\ $$$$\mathrm{0}<{I}<\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{3}+\frac{\mathrm{1}}{{e}}\right) \\ $$$$\mathrm{0}<{I}<\mathrm{1}.\mathrm{68} \\ $$$${but}\:{from}\:{attached}\:{graph}\:{the}\:{area}\:{is}\:{A}+{B} \\ $$$${but}\:{we}\:{found}\:{A}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{3}+\frac{\mathrm{1}}{{e}}\right) \\ $$$${so}\:{actual}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}+\frac{\mathrm{1}}{{e}^{{x}^{\mathrm{2}} } }\right){dx}={A}+{B} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 21/Jan/19

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