Question Number 170501 by kndramaths last updated on 25/May/22
$$\:\:\:\:\:\:\:\:\:\:{solve}\:{this}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int\int_{{D}} {x}^{\mathrm{2}} {e}^{{xy}} {dxdy} \\ $$$${D}:\left\{\left({x}.{y}\right)\in{R}^{\mathrm{2}} \:/\mathrm{0}\leqslant{x}\leqslant\mathrm{1}.\:\:\mathrm{0}\leqslant{y}\leqslant\mathrm{2}\right\} \\ $$$$\:\:\:\:\:\:\int\underset{{D}} {\int}\frac{{ydxdy}}{\left(\mathrm{1}+{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} }.\:\:\mathrm{0}\leqslant{x}\leqslant\mathrm{1}.\mathrm{0}\leqslant{y}\leqslant\mathrm{1}. \\ $$
Answered by FelipeLz last updated on 26/May/22
![A = ∫∫_D x^2 e^(xy) dxdy = ∫_0 ^2 ∫_0 ^1 x^2 e^(xy) dxdy = ∫_0 ^1 [∫_0 ^2 x^2 e^(xy) dy]dx ∫_0 ^2 x^2 e^(xy) dy = x^2 ∫_0 ^2 e^(xy) dy = [x^2 (e^(xy) /x)]_(y=0) ^(y=2) = [xe^(xy) ]_(y=0) ^(y=2) = xe^(2x) −x A = ∫_0 ^1 (xe^(2x) −x)dx = ∫_0 ^1 xe^(2x) dx−∫_0 ^1 xdx = [(1/2)xe^(2x) −(1/4)e^(2x) −(x^2 /2)]_0 ^1 = (1/2)e^2 −(1/4)e^2 −(1/2)−(0−(1/4)−0) = (1/4)(e^2 −1) • B = ∫∫_D (y/((1+x^2 +y^2 )^(3/2) ))dxdy = ∫_0 ^1 [∫(y/( (√((1+x^2 +y^2 )^3 ))))dy]_(y=0) ^(y=1) dx u = 1+x^2 +y^2 ⇒ du = 2ydy ∫(y/( (√((1+x^2 +y^2 )^3 ))))dy = (1/2)∫(1/( (√u^3 )))du = −(1/( (√u))) = −(1/( (√(1+x^2 +y^2 )))) B = ∫_0 ^1 [−(1/( (√(2+x^2 ))))+(1/( (√(1+x^2 ))))]dx = ∫_0 ^1 (1/( (√(1+x^2 ))))dx−∫_0 ^1 (1/( (√(2+x^2 ))))dx = ∫_0 ^1 (1/( (√(1+x^2 ))))dx−(1/( (√2)))∫_0 ^1 (1/( (√(1+(x^2 /2)))))dx x = tan(v) ⇒ dx = sec^2 (v)dv ∫(1/( (√(1+x^2 ))))dx = ∫((sec^2 (v))/( (√(1+tan^2 (v)))))dv = ∫sec(v)dv = ln∣sec(v)+tan(v)∣ = ln∣(√(1+x^2 ))+x∣ = sinh^(−1) (x) (x/( (√2))) = tan(t) ⇒ dx = (√2)sec^2 (t)dt (1/( (√2)))∫(1/( (√(1+(x^2 /2)))))dx = ∫((sec^2 (t))/( (√(1+tan^2 (t)))))dt = ln∣sec(t)+tan(t)∣ = ln∣(√(1+(x^2 /2)))+(x/( (√2)))∣ = sinh^(−1) ((x/( (√2)))) B = [sinh^(−1) (x)]_0 ^1 −[sinh^(−1) ((x/( (√2))))]_0 ^1 = sinh^(−1) (1)−sinh^(−1) ((1/( (√2)))) = ln((√2)+1)−ln(((1+(√3))/( (√2)))) = ln(((2+(√2))/(1+(√3))))](Q170526.png)
$${A}\:=\:\int\int_{{D}} {x}^{\mathrm{2}} {e}^{{xy}} {dxdy}\:=\:\underset{\mathrm{0}} {\overset{\mathrm{2}} {\int}}\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}{x}^{\mathrm{2}} {e}^{{xy}} {dxdy}\:=\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\left[\underset{\mathrm{0}} {\overset{\mathrm{2}} {\int}}{x}^{\mathrm{2}} {e}^{{xy}} {dy}\right]{dx} \\ $$$$\:\:\:\:\:\:\:\:\:\underset{\mathrm{0}} {\overset{\mathrm{2}} {\int}}{x}^{\mathrm{2}} {e}^{{xy}} {dy}\:=\:{x}^{\mathrm{2}} \underset{\mathrm{0}} {\overset{\mathrm{2}} {\int}}{e}^{{xy}} {dy}\:=\:\left[{x}^{\mathrm{2}} \frac{{e}^{{xy}} }{{x}}\right]_{{y}=\mathrm{0}} ^{{y}=\mathrm{2}} \:=\:\left[{xe}^{{xy}} \right]_{{y}=\mathrm{0}} ^{{y}=\mathrm{2}} \:=\:{xe}^{\mathrm{2}{x}} −{x} \\ $$$${A}\:=\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\left({xe}^{\mathrm{2}{x}} −{x}\right){dx}\:=\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}{xe}^{\mathrm{2}{x}} {dx}−\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}{xdx}\:=\:\left[\frac{\mathrm{1}}{\mathrm{2}}{xe}^{\mathrm{2}{x}} −\frac{\mathrm{1}}{\mathrm{4}}{e}^{\mathrm{2}{x}} −\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right]_{\mathrm{0}} ^{\mathrm{1}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}{e}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}{e}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}−\left(\mathrm{0}−\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{0}\right)\:=\:\frac{\mathrm{1}}{\mathrm{4}}\left({e}^{\mathrm{2}} −\mathrm{1}\right) \\ $$$$\bullet \\ $$$${B}\:=\:\int\int_{{D}} \frac{{y}}{\left(\mathrm{1}+{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)^{\mathrm{3}/\mathrm{2}} }{dxdy}\:=\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\left[\int\frac{{y}}{\:\sqrt{\left(\mathrm{1}+{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)^{\mathrm{3}} }}{dy}\right]_{{y}=\mathrm{0}} ^{{y}=\mathrm{1}} {dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:{u}\:=\:\mathrm{1}+{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \:\Rightarrow\:{du}\:=\:\mathrm{2}{ydy}\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int\frac{{y}}{\:\sqrt{\left(\mathrm{1}+{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)^{\mathrm{3}} }}{dy}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}}{\:\sqrt{{u}^{\mathrm{3}} }}{du}\:=\:−\frac{\mathrm{1}}{\:\sqrt{{u}}}\:=\:−\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }} \\ $$$${B}\:=\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\left[−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}+{x}^{\mathrm{2}} }}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\right]{dx}\:=\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{dx}−\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}+{x}^{\mathrm{2}} }}{dx}\:=\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{dx}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}}}{dx} \\ $$$$\:\:\:\:\:\:\:\:\:{x}\:=\:\mathrm{tan}\left({v}\right)\:\Rightarrow\:{dx}\:=\:\mathrm{sec}^{\mathrm{2}} \left({v}\right){dv} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{dx}\:=\:\int\frac{\mathrm{sec}^{\mathrm{2}} \left({v}\right)}{\:\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \left({v}\right)}}{dv}\:=\:\int\mathrm{sec}\left({v}\right){dv}\:=\:\mathrm{ln}\mid\mathrm{sec}\left({v}\right)+\mathrm{tan}\left({v}\right)\mid\:=\:\mathrm{ln}\mid\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }+{x}\mid\:=\:\mathrm{sinh}^{−\mathrm{1}} \left({x}\right) \\ $$$$\:\:\:\:\:\:\:\:\frac{{x}}{\:\sqrt{\mathrm{2}}}\:=\:\mathrm{tan}\left({t}\right)\:\Rightarrow\:{dx}\:=\:\sqrt{\mathrm{2}}\mathrm{sec}^{\mathrm{2}} \left({t}\right){dt} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}}}{dx}\:=\:\int\frac{\mathrm{sec}^{\mathrm{2}} \left({t}\right)}{\:\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \left({t}\right)}}{dt}\:=\:\mathrm{ln}\mid\mathrm{sec}\left({t}\right)+\mathrm{tan}\left({t}\right)\mid\:=\:\mathrm{ln}\mid\sqrt{\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}}+\frac{{x}}{\:\sqrt{\mathrm{2}}}\mid\:=\:\mathrm{sinh}^{−\mathrm{1}} \left(\frac{{x}}{\:\sqrt{\mathrm{2}}}\right) \\ $$$${B}\:=\:\left[\mathrm{sinh}^{−\mathrm{1}} \left({x}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} −\left[\mathrm{sinh}^{−\mathrm{1}} \left(\frac{{x}}{\:\sqrt{\mathrm{2}}}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \:=\:\mathrm{sinh}^{−\mathrm{1}} \left(\mathrm{1}\right)−\mathrm{sinh}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)\:=\:\mathrm{ln}\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)−\mathrm{ln}\left(\frac{\mathrm{1}+\sqrt{\mathrm{3}}}{\:\sqrt{\mathrm{2}}}\right)\:=\:\mathrm{ln}\left(\frac{\mathrm{2}+\sqrt{\mathrm{2}}}{\mathrm{1}+\sqrt{\mathrm{3}}}\right) \\ $$