Question Number 66550 by Tony Lin last updated on 17/Aug/19
$$\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\frac{{y}}{\mathrm{2}}\:} ^{\frac{\mathrm{1}}{\mathrm{2}}\:} {e}^{−{x}^{\mathrm{2}} } {dxdy}=? \\ $$
Commented by ~ À ® @ 237 ~ last updated on 17/Aug/19
![let named it I I=∫∫_D e^(−x^2 ) dxdy with D={(x,y) / 0<y<1 , (y/2)<x<(1/2) } we ought to transform that domain , in the way to start the integration on y ( if not it will be on x ) D={(x,y) / 0<x<(1/2) , 0<y<2x } ( you can see it better with a graph) Now I=∫_0 ^(1/2) (∫_0 ^(2x) e^(−x^2 ) dy) dx = ∫_0 ^(1/2) [ye^(−x^2 ) ]_0 ^(2x) dx=∫_0_ ^(1/2) 2xe^(−x^2 ) dx =[−e^(−x^2 ) ]_0 ^(1/2) =−e^(−(1/4)) +1 So I=1− e^((−1)/4)](Q66570.png)
$${let}\:{named}\:{it}\:{I} \\ $$$${I}=\int\int_{{D}} {e}^{−{x}^{\mathrm{2}} } {dxdy}\:\:\:\:{with}\:\:{D}=\left\{\left({x},{y}\right)\:\:/\:\:\mathrm{0}<{y}<\mathrm{1}\:\:,\:\frac{{y}}{\mathrm{2}}<{x}<\frac{\mathrm{1}}{\mathrm{2}}\:\right\} \\ $$$${we}\:{ought}\:{to}\:{transform}\:{that}\:{domain}\:,\:{in}\:{the}\:{way}\:{to}\:{start}\:{the}\:{integration}\:{on}\:{y}\:\left(\:{if}\:{not}\:{it}\:{will}\:{be}\:{on}\:{x}\:\right) \\ $$$${D}=\left\{\left({x},{y}\right)\:/\:\:\mathrm{0}<{x}<\frac{\mathrm{1}}{\mathrm{2}}\:\:,\:\:\mathrm{0}<{y}<\mathrm{2}{x}\:\right\}\:\left(\:{you}\:{can}\:{see}\:{it}\:{better}\:{with}\:{a}\:{graph}\right) \\ $$$${Now}\: \\ $$$${I}=\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \left(\int_{\mathrm{0}} ^{\mathrm{2}{x}} \:{e}^{−{x}^{\mathrm{2}} } \:{dy}\right)\:{dx}\:=\:\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \left[{ye}^{−{x}^{\mathrm{2}} } \right]_{\mathrm{0}} ^{\mathrm{2}{x}} {dx}=\int_{\mathrm{0}_{} } ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\mathrm{2}{xe}^{−{x}^{\mathrm{2}} } {dx}\:=\left[−{e}^{−{x}^{\mathrm{2}} } \right]_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:=−{e}^{−\frac{\mathrm{1}}{\mathrm{4}}} +\mathrm{1} \\ $$$$\:\:{So}\:\:\:{I}=\mathrm{1}−\:{e}^{\frac{−\mathrm{1}}{\mathrm{4}}} \\ $$
Commented by Tony Lin last updated on 17/Aug/19
$${thanks}\:{sir} \\ $$