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Question Number 190487 by stvnmaxi last updated on 03/Apr/23

Answered by aleks041103 last updated on 04/Apr/23

A=∫_0 ^1 ∫_0 ^1 x^2 y^2 dxdy =(∫_0 ^1 x^2 dx)(∫_0 ^1 y^2 dy)= =(1/3) (1/3)=(1/9)=A B=∫_0 ^1 ∫_0 ^1 (x^2 +y^2 )dxdy= =∫_0 ^1 ∫_0 ^1 x^2 dxdy + ∫_0 ^1 ∫_0 ^1 y^2 dxdy= =(∫_0 ^1 dy)(∫_0 ^1 x^2 dx)+(∫_0 ^1 dx)(∫_0 ^1 y^2 dy)= =1×(1/3)+1×(1/3)=(2/3)=B C=∫_1 ^(√3) ∫_0 ^1 ((dxdy)/(x^2 +y^2 ))= =∫_1 ^(√3) (∫_0 ^1 (dx/(x^2 +y^2 )))dy ∫_0 ^1 (dx/(x^2 +y^2 ))=(1/y)∫_0 ^(1/y) ((d(x/y))/((x/y)^2 +1))=(1/y)arctan((1/y^2 )) ⇒C=∫_1 ^(√3) arctan((1/y^2 ))(dy/y)= =(1/2)∫_1 ^(√3) arctan(1/y^2 )((d(y^2 ))/y^2 )= =(1/2)∫_1 ^3 ((arctan(1/z))/z)dz this integral is not solvable in elementary functions

$${A}=\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{2}} {y}^{\mathrm{2}} {dxdy}\:=\left(\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{2}} {dx}\right)\left(\int_{\mathrm{0}} ^{\mathrm{1}} {y}^{\mathrm{2}} {dy}\right)= \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\:\frac{\mathrm{1}}{\mathrm{3}}=\frac{\mathrm{1}}{\mathrm{9}}={A} \\ $$$${B}=\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} \left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right){dxdy}= \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{2}} {dxdy}\:+\:\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} {y}^{\mathrm{2}} {dxdy}= \\ $$$$=\left(\int_{\mathrm{0}} ^{\mathrm{1}} {dy}\right)\left(\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{2}} {dx}\right)+\left(\int_{\mathrm{0}} ^{\mathrm{1}} {dx}\right)\left(\int_{\mathrm{0}} ^{\mathrm{1}} {y}^{\mathrm{2}} {dy}\right)= \\ $$$$=\mathrm{1}×\frac{\mathrm{1}}{\mathrm{3}}+\mathrm{1}×\frac{\mathrm{1}}{\mathrm{3}}=\frac{\mathrm{2}}{\mathrm{3}}={B} \\ $$$${C}=\int_{\mathrm{1}} ^{\sqrt{\mathrm{3}}} \int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dxdy}}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }= \\ $$$$=\int_{\mathrm{1}} ^{\sqrt{\mathrm{3}}} \left(\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dx}}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\right){dy} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dx}}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }=\frac{\mathrm{1}}{{y}}\int_{\mathrm{0}} ^{\mathrm{1}/{y}} \frac{{d}\left({x}/{y}\right)}{\left({x}/{y}\right)^{\mathrm{2}} +\mathrm{1}}=\frac{\mathrm{1}}{{y}}{arctan}\left(\frac{\mathrm{1}}{{y}^{\mathrm{2}} }\right) \\ $$$$\Rightarrow{C}=\int_{\mathrm{1}} ^{\sqrt{\mathrm{3}}} {arctan}\left(\frac{\mathrm{1}}{{y}^{\mathrm{2}} }\right)\frac{{dy}}{{y}}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{1}} ^{\sqrt{\mathrm{3}}} {arctan}\left(\mathrm{1}/{y}^{\mathrm{2}} \right)\frac{{d}\left({y}^{\mathrm{2}} \right)}{{y}^{\mathrm{2}} }= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{1}} ^{\mathrm{3}} \frac{{arctan}\left(\mathrm{1}/{z}\right)}{{z}}{dz} \\ $$$${this}\:{integral}\:{is}\:{not}\:{solvable}\:{in}\:{elementary} \\ $$$${functions} \\ $$

Commented by stvnmaxi last updated on 05/Apr/23

there is not another way to calculer the last question (C)

$$\mathrm{there}\:\mathrm{is}\:\mathrm{not}\:\mathrm{another}\:\mathrm{way}\:\mathrm{to}\:\mathrm{calculer}\:\mathrm{the}\:\mathrm{last}\:\mathrm{question}\:\left(\mathrm{C}\right) \\ $$

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