Question Number 31084 by abdo imad last updated on 02/Mar/18
$${find}\:\int\int_{{D}} \:\:\frac{{dxdy}}{\left({x}+{y}\right)^{\mathrm{4}} }\:\:{with}\:{D}=\left\{\left({x},{y}\right)\in{R}^{\mathrm{2}} /{x}\geqslant\mathrm{1},{y}\geqslant\mathrm{1},{x}+{y}\leqslant\mathrm{4}\right\} \\ $$
Commented by abdo imad last updated on 11/Mar/18
![x+y ≤4 and y≥1 ⇒−y≤−1 but x ≤4−y ⇒1≤x≤3 and 1≤y≤4−x I =∫_1 ^3 (∫_1 ^(4−x) (dy/((x+y)^4 )))dx but ∫_1 ^(4−x) (x+y)^(−4) dy=[ (1/(−3))(x+y)^(−3) ]_(y=1) ^(y=4−x) =(1/3)(x+1)^(−3) −(1/3)(4)^(−3) =(1/3)(x+1)^(−3) −(1/(3.4^3 )) I=∫_1 ^3 ((1/3)(x+1)^(−3) −(1/(192)))dx=((−2)/(192)) +(1/3)[(1/(−2))(x+1)^(−2) ]_1 ^3 =−(1/(96)) −(1/6)(4^(−2) −2^(−2) )=−(1/(96)) −(1/6)((1/(16)) −(1/4)) =−(1/6) ((−3)/(16)) −(1/(96)) = (1/(32)) −(1/(96)) .](https://www.tinkutara.com/question/Q31624.png)
$${x}+{y}\:\leqslant\mathrm{4}\:\:{and}\:{y}\geqslant\mathrm{1}\:\Rightarrow−{y}\leqslant−\mathrm{1}\:{but}\:{x}\:\leqslant\mathrm{4}−{y}\:\Rightarrow\mathrm{1}\leqslant{x}\leqslant\mathrm{3}\:{and} \\ $$$$\mathrm{1}\leqslant{y}\leqslant\mathrm{4}−{x} \\ $$$${I}\:=\int_{\mathrm{1}} ^{\mathrm{3}} \:\left(\int_{\mathrm{1}} ^{\mathrm{4}−{x}} \:\:\:\frac{{dy}}{\left({x}+{y}\right)^{\mathrm{4}} }\right){dx}\:\:{but} \\ $$$$\int_{\mathrm{1}} ^{\mathrm{4}−{x}} \:\left({x}+{y}\right)^{−\mathrm{4}} {dy}=\left[\:\frac{\mathrm{1}}{−\mathrm{3}}\left({x}+{y}\right)^{−\mathrm{3}} \right]_{{y}=\mathrm{1}} ^{{y}=\mathrm{4}−{x}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\left({x}+\mathrm{1}\right)^{−\mathrm{3}} \:−\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{4}\right)^{−\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{3}}\left({x}+\mathrm{1}\right)^{−\mathrm{3}} \:−\frac{\mathrm{1}}{\mathrm{3}.\mathrm{4}^{\mathrm{3}} } \\ $$$${I}=\int_{\mathrm{1}} ^{\mathrm{3}} \left(\frac{\mathrm{1}}{\mathrm{3}}\left({x}+\mathrm{1}\right)^{−\mathrm{3}} \:−\frac{\mathrm{1}}{\mathrm{192}}\right){dx}=\frac{−\mathrm{2}}{\mathrm{192}}\:+\frac{\mathrm{1}}{\mathrm{3}}\left[\frac{\mathrm{1}}{−\mathrm{2}}\left({x}+\mathrm{1}\right)^{−\mathrm{2}} \right]_{\mathrm{1}} ^{\mathrm{3}} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{96}}\:−\frac{\mathrm{1}}{\mathrm{6}}\left(\mathrm{4}^{−\mathrm{2}} \:−\mathrm{2}^{−\mathrm{2}} \right)=−\frac{\mathrm{1}}{\mathrm{96}}\:−\frac{\mathrm{1}}{\mathrm{6}}\left(\frac{\mathrm{1}}{\mathrm{16}}\:−\frac{\mathrm{1}}{\mathrm{4}}\right) \\ $$$$=−\frac{\mathrm{1}}{\mathrm{6}}\:\frac{−\mathrm{3}}{\mathrm{16}}\:−\frac{\mathrm{1}}{\mathrm{96}}\:=\:\frac{\mathrm{1}}{\mathrm{32}}\:−\frac{\mathrm{1}}{\mathrm{96}}\:. \\ $$