∫_0 ^1 ∫_((y/2) ) ^((1/2) ) e^(−x^2 ) dxdy=?


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 66550 by Tony Lin last updated on 17/Aug/19

$\int_{0}^{1} \int_{\frac{y}{2}}^{\frac{1}{2}} e^{- x^{2}} d x d y = ?$
Commented by ~ À ® @ 237 ~ last updated on 17/Aug/19
$let named it I I=∫∫_D e^(−x^2 ) dxdy with D={(x,y) / 0<y<1 , (y/2)<x<(1/2) } we ought to transform that domain , in the way to start the integration on y ( if not it will be on x ) D={(x,y) / 0<x<(1/2) , 0<y<2x } ( you can see it better with a graph) Now I=∫_0 ^(1/2) (∫_0 ^(2x) e^(−x^2 ) dy) dx = ∫_0 ^(1/2) [ye^(−x^2 ) ]_0 ^(2x) dx=∫_0_ ^(1/2) 2xe^(−x^2 ) dx =[−e^(−x^2 ) ]_0 ^(1/2) =−e^(−(1/4)) +1 So I=1− e^((−1)/4)$
$l e t n a m e d i t I$ $I = \int \int_{D} e^{- x^{2}} d x d y w i t h D = {(x, y) / 0 < y < 1, \frac{y}{2} < x < \frac{1}{2}}$ $w e o u g h t t o t r a n s f o r m t h a t d o m a i n, i n t h e w a y t o s t a r t t h e i n t e g r a t i o n o n y (i f n o t i t w i l l b e o n x)$ $D = {(x, y) / 0 < x < \frac{1}{2}, 0 < y < 2 x} (y o u c a n s e e i t b e t t e r w i t h a g r a p h)$ $N o w$ $I = \int_{0}^{\frac{1}{2}} (\int_{0}^{2 x} e^{- x^{2}} d y) d x = \int_{0}^{\frac{1}{2}} {[{y e}^{- x^{2}}]}_{0}^{2 x} d x = \int_{0_{}}^{\frac{1}{2}} 2 {x e}^{- x^{2}} d x = {[- e^{- x^{2}}]}_{0}^{\frac{1}{2}} = - e^{- \frac{1}{4}} + 1$ $S o I = 1 - e^{\frac{- 1}{4}}$
Commented by Tony Lin last updated on 17/Aug/19

$t h a n k s s i r$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum