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Question Number 47182 by maxmathsup by imad last updated on 05/Nov/18

$$calculate{\int }_0^1{e}^{-x}\sqrt{1-\sqrt{x}}dx$$

Commented by tanmay.chaudhury50@gmail.com last updated on 07/Nov/18

Commented by tanmay.chaudhury50@gmail.com last updated on 07/Nov/18

$$graphof{e}^{-x}\sqrt{1-\sqrt{x}}$$$$0.50>{\int }_0^1{e}^{-x}\sqrt{1-\sqrt{x}}dx>0$$

Commented by maxmathsup by imad last updated on 11/Nov/18

$$letdetermineaapproximatvalueofA={\int }_0^1{e}^{-x}\sqrt{1-\sqrt{x}}dxwehave$$$$e^u=1+\frac{u}{1!}+\frac{u^2}{2!}+....\Rightarrow e^{-x}=1-x+\frac{x^2}{2}-...\Rightarrow 0⩽e^{-x}⩽1-x+\frac{x^2}{2}\Rightarrow$$$$0⩽e^{-x}\sqrt{1-\sqrt{x}}⩽(1-x+\frac{x^2}{2})\sqrt{1-\sqrt{x}}\Rightarrow 0⩽{\int }_0^1{e}^{-x}\sqrt{1-\sqrt{x}}dx⩽{\int }_0^1(1-x+\frac{x^2}{2})\sqrt{1-\sqrt{x}}dx$$$$changement\sqrt{1-\sqrt{x}}=tgive1-\sqrt{x}=t^2\Rightarrow \sqrt{x}=1-t^2\Rightarrow x={(1-t^2)}^2=t^4-2t^2+1\Rightarrow$$$${\int }_0^1(1-x+\frac{x^2}{2})\sqrt{1-\sqrt{x}}dx=-{\int }_0^1(1-t^4+2t^2-1+\frac{{(t^4-2t^2+1)}^2}{2})t(4t^3-4t)dt$$$$=-2{\int }_0^1(-2t^4+4t^2+(t^8+4t^4+1+2(-2t^6+t^4-2t^2)(t^4-t^2)dt$$$$=-2{\int }_0^1(-2t^4+4t^2+t^8+6t^4-4t^6-4t^2+1)(t^4-t^2)dt$$$$=-2{\int }_0^1(t^8-4t^6-2t^4-4t^2+1)(t^4-t^2)dt$$$$=-2{\int }_0^1(t^{12}-t^{10}-4t^{10}+4t^8-2t^8+2t^6-4t^6+4t^4)dt$$$$=-2{\int }_0^1(t^{12}-5t^{10}+2t^8-2t^6+4t^4)dt$$$$=-2{[\frac{t^{13}}{13}-\frac{5t^{11}}{11}+\frac{2t^9}{9}-\frac{2t^7}{7}+\frac{4t^5}{5}]}_0^1$$$$=-2(\frac{1}{13}-\frac{5}{11}+\frac{2}{9}-\frac{2}{7}+\frac{4}{5})=-\frac{2}{13}+\frac{10}{11}-\frac{4}{9}+\frac{4}{7}-\frac{8}{5}={\alpha }_0\Rightarrow 0<A⩽{\alpha }_0$$

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