Question Number 65401 by mathmax by abdo last updated on 29/Jul/19
$${let}\:{f}\left({x},{y}\right)=\left({x}+{y}\right)\sqrt{{x}+{y}−\mathrm{1}} \\ $$$${calculate}\:\:\int\int_{{D}} {f}\left({x},{y}\right){dxdy}\:{with}\: \\ $$$${D}\:=\left\{\left({x},{y}\right)\in{R}^{\mathrm{2}} /\:\:\mathrm{1}\leqslant{x}\leqslant\mathrm{2}\:\:{and}\:\:\:\mathrm{1}\leqslant{y}\leqslant\sqrt{\mathrm{3}}\right\} \\ $$
Commented by mathmax by abdo last updated on 01/Aug/19
![∫∫_D f(x,y)dxdy =∫_1 ^(√3) (∫_1 ^2 (x+y)(√(x+y−1))dx)dy=∫_1 ^(√3) A(y)dy changement (√(x+y−1))=t give x+y−1=t^2 ⇒x =t^2 −y+1 ⇒ A(y) =∫_(√y) ^(√(y+1)) (t^2 −y+1+y)t(2t)dt =2 ∫_(√y) ^(√(y+1)) (t^4 +t^2 )dt =2[(t^5 /5)+(t^3 /3)]_(√y) ^(√(y+1)) =2{((((√(y+1)))^5 )/5)+((((√(y+1)))^3 )/3) −((((√y))^5 )/5)−((((√y))^3 )/3) =2{ (((y+1)^2 (√(y+1)))/5) +(((y+1)(√(y+1)))/3)−((y^2 (√y))/5)−((y(√y))/3)} ⇒ ∫∫_D f(x,y)dxdy =(2/5)∫_1 ^(√3) (y+1)^2 (√(y+1)) dy +(2/3)∫_1 ^(√3) (y+1)(√(y+1))dy −(2/5) ∫_1 ^(√3) y^2 (√y)dy −(2/3) ∫_1 ^(√3) y(√y)dy changement (√(y+1))=u give y+1 =u^2 ⇒ ∫_1 ^(√3) (y+1)^2 (√(y+1))dy =∫_(√2) ^(√((√3)+1)) u^4 u (2u)du =2 ∫_(√2) ^(√((√3)+1)) u^6 du =(2/7)[u^7 ]_(√2) ^(√((√3)+1)) =(2/7){ ((√((√3)+1)))^7 −((√2))^7 } ∫_1 ^(√3) (y+1)(√(y+1))dy =∫_(√2) ^(√((√3)+1)) u^2 u(2udu) =2 ∫_(√2) ^(√((√3)+1)) u^4 du =(2/5){((√((√3)+1)))^5 −((√2))^5 } also changement (√y)=u give y=u^2 ⇒∫_1 ^(√3) y^2 (√y)dy =∫_1 ^(√(√3)) u^4 (u)(2udu) =2 ∫_1 ^(√(√3)) u^6 du =(2/7)[((√(√3)))^7 −1) ∫_1 ^(√3) y(√y)dy =∫_1 ^(√(√3)) u^2 u(2u)du =2 ∫_1 ^(√(√3)) u^4 xu =(2/5){ ((√(√3)))^5 −1} the value of this integral is known](https://www.tinkutara.com/question/Q65657.png)
$$\int\int_{{D}} {f}\left({x},{y}\right){dxdy}\:=\int_{\mathrm{1}} ^{\sqrt{\mathrm{3}}} \left(\int_{\mathrm{1}} ^{\mathrm{2}} \left({x}+{y}\right)\sqrt{{x}+{y}−\mathrm{1}}{dx}\right){dy}=\int_{\mathrm{1}} ^{\sqrt{\mathrm{3}}} {A}\left({y}\right){dy} \\ $$$${changement}\:\sqrt{{x}+{y}−\mathrm{1}}={t}\:{give}\:{x}+{y}−\mathrm{1}={t}^{\mathrm{2}} \:\Rightarrow{x}\:={t}^{\mathrm{2}} −{y}+\mathrm{1}\:\Rightarrow \\ $$$${A}\left({y}\right)\:=\int_{\sqrt{{y}}} ^{\sqrt{{y}+\mathrm{1}}} \left({t}^{\mathrm{2}} −{y}+\mathrm{1}+{y}\right){t}\left(\mathrm{2}{t}\right){dt} \\ $$$$=\mathrm{2}\:\int_{\sqrt{{y}}} ^{\sqrt{{y}+\mathrm{1}}} \left({t}^{\mathrm{4}} +{t}^{\mathrm{2}} \right){dt}\:=\mathrm{2}\left[\frac{{t}^{\mathrm{5}} }{\mathrm{5}}+\frac{{t}^{\mathrm{3}} }{\mathrm{3}}\right]_{\sqrt{{y}}} ^{\sqrt{{y}+\mathrm{1}}} \:=\mathrm{2}\left\{\frac{\left(\sqrt{{y}+\mathrm{1}}\right)^{\mathrm{5}} }{\mathrm{5}}+\frac{\left(\sqrt{{y}+\mathrm{1}}\right)^{\mathrm{3}} }{\mathrm{3}}\right. \\ $$$$−\frac{\left(\sqrt{{y}}\right)^{\mathrm{5}} }{\mathrm{5}}−\frac{\left(\sqrt{{y}}\right)^{\mathrm{3}} }{\mathrm{3}}\:=\mathrm{2}\left\{\:\frac{\left({y}+\mathrm{1}\right)^{\mathrm{2}} \sqrt{{y}+\mathrm{1}}}{\mathrm{5}}\:+\frac{\left({y}+\mathrm{1}\right)\sqrt{{y}+\mathrm{1}}}{\mathrm{3}}−\frac{{y}^{\mathrm{2}} \sqrt{{y}}}{\mathrm{5}}−\frac{{y}\sqrt{{y}}}{\mathrm{3}}\right\}\:\Rightarrow \\ $$$$\int\int_{{D}} {f}\left({x},{y}\right){dxdy}\:=\frac{\mathrm{2}}{\mathrm{5}}\int_{\mathrm{1}} ^{\sqrt{\mathrm{3}}} \left({y}+\mathrm{1}\right)^{\mathrm{2}} \sqrt{{y}+\mathrm{1}}\:{dy}\:+\frac{\mathrm{2}}{\mathrm{3}}\int_{\mathrm{1}} ^{\sqrt{\mathrm{3}}} \left({y}+\mathrm{1}\right)\sqrt{{y}+\mathrm{1}}{dy} \\ $$$$−\frac{\mathrm{2}}{\mathrm{5}}\:\int_{\mathrm{1}} ^{\sqrt{\mathrm{3}}} {y}^{\mathrm{2}} \sqrt{{y}}{dy}\:\:−\frac{\mathrm{2}}{\mathrm{3}}\:\int_{\mathrm{1}} ^{\sqrt{\mathrm{3}}} {y}\sqrt{{y}}{dy}\: \\ $$$${changement}\:\sqrt{{y}+\mathrm{1}}={u}\:{give}\:{y}+\mathrm{1}\:={u}^{\mathrm{2}} \:\Rightarrow \\ $$$$\int_{\mathrm{1}} ^{\sqrt{\mathrm{3}}} \left({y}+\mathrm{1}\right)^{\mathrm{2}} \sqrt{{y}+\mathrm{1}}{dy}\:=\int_{\sqrt{\mathrm{2}}} ^{\sqrt{\sqrt{\mathrm{3}}+\mathrm{1}}} {u}^{\mathrm{4}} {u}\:\left(\mathrm{2}{u}\right){du}\:=\mathrm{2}\:\int_{\sqrt{\mathrm{2}}} ^{\sqrt{\sqrt{\mathrm{3}}+\mathrm{1}}} {u}^{\mathrm{6}} {du}\: \\ $$$$=\frac{\mathrm{2}}{\mathrm{7}}\left[{u}^{\mathrm{7}} \right]_{\sqrt{\mathrm{2}}} ^{\sqrt{\sqrt{\mathrm{3}}+\mathrm{1}}} \:=\frac{\mathrm{2}}{\mathrm{7}}\left\{\:\:\left(\sqrt{\sqrt{\mathrm{3}}+\mathrm{1}}\right)^{\mathrm{7}} −\left(\sqrt{\mathrm{2}}\right)^{\mathrm{7}} \right\} \\ $$$$\int_{\mathrm{1}} ^{\sqrt{\mathrm{3}}} \left({y}+\mathrm{1}\right)\sqrt{{y}+\mathrm{1}}{dy}\:=\int_{\sqrt{\mathrm{2}}} ^{\sqrt{\sqrt{\mathrm{3}}+\mathrm{1}}} {u}^{\mathrm{2}} {u}\left(\mathrm{2}{udu}\right)\:=\mathrm{2}\:\int_{\sqrt{\mathrm{2}}} ^{\sqrt{\sqrt{\mathrm{3}}+\mathrm{1}}} \:\:{u}^{\mathrm{4}} {du} \\ $$$$=\frac{\mathrm{2}}{\mathrm{5}}\left\{\left(\sqrt{\sqrt{\mathrm{3}}+\mathrm{1}}\right)^{\mathrm{5}} −\left(\sqrt{\mathrm{2}}\right)^{\mathrm{5}} \right\}\:\:{also}\:{changement}\:\sqrt{{y}}={u}\:{give}\:{y}={u}^{\mathrm{2}} \\ $$$$\Rightarrow\int_{\mathrm{1}} ^{\sqrt{\mathrm{3}}} {y}^{\mathrm{2}} \sqrt{{y}}{dy}\:=\int_{\mathrm{1}} ^{\sqrt{\sqrt{\mathrm{3}}}} {u}^{\mathrm{4}} \left({u}\right)\left(\mathrm{2}{udu}\right)\:=\mathrm{2}\:\int_{\mathrm{1}} ^{\sqrt{\sqrt{\mathrm{3}}}} {u}^{\mathrm{6}} {du}\:=\frac{\mathrm{2}}{\mathrm{7}}\left[\left(\sqrt{\sqrt{\mathrm{3}}}\right)^{\mathrm{7}} −\mathrm{1}\right) \\ $$$$\int_{\mathrm{1}} ^{\sqrt{\mathrm{3}}} {y}\sqrt{{y}}{dy}\:=\int_{\mathrm{1}} ^{\sqrt{\sqrt{\mathrm{3}}}} \:{u}^{\mathrm{2}} {u}\left(\mathrm{2}{u}\right){du}\:=\mathrm{2}\:\int_{\mathrm{1}} ^{\sqrt{\sqrt{\mathrm{3}}}} {u}^{\mathrm{4}} {xu}\:=\frac{\mathrm{2}}{\mathrm{5}}\left\{\:\left(\sqrt{\sqrt{\mathrm{3}}}\right)^{\mathrm{5}} −\mathrm{1}\right\} \\ $$$${the}\:{value}\:{of}\:{this}\:{integral}\:{is}\:{known} \\ $$