∫_1 ^2 ∫_1 ^2 ln(x+y)dx dy


Question and Answers Forum

All Questions Topic List
Geometry Questions
Previous in All Question Next in All Question
Previous in Geometry Next in Geometry
Question Number 24831 by Eng.Firas last updated on 27/Nov/17

$\int_{1}^{2} \int_{1}^{2} l n (x + y) d x d y$
	Answered by prakash jain last updated on 27/Nov/17
	$∫_1 ^2 ∫_1 ^2 ln(x+y)dxdy =∫_1 ^2 [(x+y)ln (x+y)−x]_1 ^2 dy =∫_1 ^2 {(2+y)ln (y+2)−2−(1+y)ln (1+y)+1}dy =∫_1 ^2 {(2+y)ln (y+2)−(1+y)ln (1+y)−1}dy ∫_1 ^2 (2+y)ln (y+2)dy =[ln (y+2)(((2+y)^2 )/2)−(((2+y)^2 )/4)]_1 ^2 =[(ln 4)(4^2 /2)−(4^2 /4)−(ln 3)(3^2 /2)+(3^2 /4)] =8ln 4−8−(9/2)ln 3+(9/4) ∫_1 ^2 (1+y)ln (y+1)dy =[ln (y+1)(((1+y)^2 )/2)−(((1+y)^2 )/4)]_1 ^2 =(9/2)ln 3−(9/4)−(2^2 /2)ln 2+(2^2 /4) ∫_1 ^2 (−1)dy=−1 ans=8ln 4−8−2ln 2+1−1 =16ln 2−8−2ln 2 =14ln 2−8$
	$\int_{1}^{2} \int_{1}^{2} l n (x + y) d x d y$ $= \int_{1}^{2} {[(x + y) \ln (x + y) - x]}_{1}^{2} d y$ $= \int_{1}^{2} {(2 + y) \ln (y + 2) - 2 - (1 + y) \ln (1 + y) + 1} d y$ $= \int_{1}^{2} {(2 + y) \ln (y + 2) - (1 + y) \ln (1 + y) - 1} d y$ $\int_{1}^{2} (2 + y) \ln (y + 2) d y$ $= {[\ln (y + 2) \frac{{(2 + y)}^{2}}{2} - \frac{{(2 + y)}^{2}}{4}]}_{1}^{2}$ $= [(\ln 4) \frac{4^{2}}{2} - \frac{4^{2}}{4} - (\ln 3) \frac{3^{2}}{2} + \frac{3^{2}}{4}]$ $= 8 \ln 4 - 8 - \frac{9}{2} \ln 3 + \frac{9}{4}$ $\int_{1}^{2} (1 + y) \ln (y + 1) d y$ $= {[\ln (y + 1) \frac{{(1 + y)}^{2}}{2} - \frac{{(1 + y)}^{2}}{4}]}_{1}^{2}$ $= \frac{9}{2} \ln 3 - \frac{9}{4} - \frac{2^{2}}{2} \ln 2 + \frac{2^{2}}{4}$ $\int_{1}^{2} (- 1) d y = - 1$ $a n s = 8 \ln 4 - 8 - 2 \ln 2 + 1 - 1$ $= 16 \ln 2 - 8 - 2 \ln 2$ $= 14 \ln 2 - 8$
	Commented by Eng.Firas last updated on 27/Nov/17

	$t h a n k y o u$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum