Calculate, using cartesian coodinates, the following integrals: 1) ∫∫_D dxdy being D={ (x,y)∈R^2 /0≤x≤(1/2),y+x≤1,y≥0} 2) ∫∫_D x^3 ydxdy being D={(x,y)∈R^2 /0≤x≤(1/2),y+x≤1,y≥0} 3) ∫∫_D (x/y)dxdy being D={(x,y)∈R^2 /xy≤16,x≥y,x−6≤y,x≥0,y≥1} Help please!


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Question Number 61258 by cesar.marval.larez@gmail.com last updated on 31/May/19
$Calculate, using cartesian coodinates, the following integrals: 1) ∫∫_D dxdy being D={ (x,y)∈R^2 /0≤x≤(1/2),y+x≤1,y≥0} 2) ∫∫_D x^3 ydxdy being D={(x,y)∈R^2 /0≤x≤(1/2),y+x≤1,y≥0} 3) ∫∫_D (x/y)dxdy being D={(x,y)∈R^2 /xy≤16,x≥y,x−6≤y,x≥0,y≥1} Help please!$
$C a l c u l a t e, u s i n g c a r t e s i a n c o o d i n a t e s, t h e f o l l o w i n g$ $i n t e g r a l s :$ $1) \int \int_{D} d x d y b e i n g D = {(x, y) \in R^{2} / 0 ⩽ x ⩽ \frac{1}{2}, y + x ⩽ 1, y ⩾ 0}$ $2) \int \int_{D} x^{3} y d x d y b e i n g D = {(x, y) \in R^{2} / 0 ⩽ x ⩽ \frac{1}{2}, y + x ⩽ 1, y ⩾ 0}$ $3) \int \int_{D} \frac{x}{y} d x d y b e i n g D = {(x, y) \in R^{2} / x y ⩽ 16, x ⩾ y, x - 6 ⩽ y, x ⩾ 0, y ⩾ 1}$ $H e l p p l e a s e!$
Commented by abdo mathsup 649 cc last updated on 31/May/19

$1) \int \int d x d y = \int_{0}^{\frac{1}{2}} (\int_{0}^{1 - x} d y) d x$ $= \int_{0}^{\frac{1}{2}} (1 - x) d x = {[x - \frac{x^{2}}{2}]}_{0}^{\frac{1}{2}} = \frac{1}{2} - \frac{1}{8} = \frac{4 - 1}{8} = \frac{3}{8} .$
Commented by abdo mathsup 649 cc last updated on 31/May/19
$2) ∫∫ x^3 y dxdy =∫_0 ^(1/2) (∫_0 ^(1−x) ydy)x^3 dx =∫_0 ^(1/2) ([(y^2 /2)]_0 ^(1−x) )x^3 dx =(1/2) ∫_0 ^(1/2) (x−1)^2 x^3 dx =(1/2) ∫_0 ^(1/2) (x^2 −2x+1)x^3 dx =(1/2) ∫_0 ^(1/2) (x^5 −2x^4 +x^3 )dx =(1/2)[(x^6 /6) −(2/5) x^5 +(x^4 /4)]_0 ^(1/2) =(1/2){ (1/(6.2^6 )) −(2/(5.2^5 )) +(1/2^4 )} =....$
$2) \int \int x^{3} y d x d y = \int_{0}^{\frac{1}{2}} (\int_{0}^{1 - x} y d y) x^{3} d x$ $= \int_{0}^{\frac{1}{2}} ({[\frac{y^{2}}{2}]}_{0}^{1 - x}) x^{3} d x = \frac{1}{2} \int_{0}^{\frac{1}{2}} {(x - 1)}^{2} x^{3} d x$ $= \frac{1}{2} \int_{0}^{\frac{1}{2}} (x^{2} - 2 x + 1) x^{3} d x$ $= \frac{1}{2} \int_{0}^{\frac{1}{2}} (x^{5} - 2 x^{4} + x^{3}) d x$ $= \frac{1}{2} {[\frac{x^{6}}{6} - \frac{2}{5} x^{5} + \frac{x^{4}}{4}]}_{0}^{\frac{1}{2}}$ $= \frac{1}{2} {\frac{1}{6 . 2^{6}} - \frac{2}{5 . 2^{5}} + \frac{1}{2^{4}}} = . . . .$
	Answered by perlman last updated on 31/May/19

	$3) D = [(x, y) . x y < 16, x > y . x - 6 < y . x > 0 . y > 1]$ $= D_{1} [(x, y) . 1 < y < x . 1 < x < 4] \cup D_{2} [(x . y) 1 < y < \frac{16}{x} . 4 < x < 3 + \sqrt{(30)}]$ $\int \int_{D 1 \cup D 2} f (x, y) d x d y = \int \int_{D 1} f (x, y) d x d y + \int \int_{D 2} f (x, y) d x d y i f D 1 \cap D 2 = \emptyset$ $m o r g e n e r a l y i s = i f \int \int_{D 1 \cap D 2} f (x, y) d x d y = 0 ‘ ‘ t h e o r i e o f {i n t e g r a t i o n}^{″}$ $S o w e i n t e g r a t e o v e r D 1 + i n t e g r a t i o n o v e r D_{2}$ $o v e r D_{1} = \int_{1}^{4} (\int_{1}^{x} \frac{x}{y} d y) d x = \int_{1}^{4} {[x l n (y)]}_{1}^{x} d x = \int_{1}^{4} (x l n (x)) d x$ $= {[\frac{x^{2}}{2} l n (x)]}_{1}^{4} - \int_{1}^{4} \frac{x}{2} d x = 2 l n (4) - {[\frac{x^{2}}{4}]}_{1}^{4} = 2 l n (4) - 4 + \frac{1}{4} = 2 l n (4) - \frac{15}{4}$ $\int_{4}^{3 + \sqrt{(30)}} \int_{1}^{\frac{16}{x}} \frac{x}{y} d y d x = \int_{4}^{3 + \sqrt{(30)}} x l n (\frac{16}{x}) d x = \int_{4}^{3 + \sqrt{(30)}} x l n (16) d x - \int_{4}^{3 + \sqrt{(30)}} x l n (x) d x$ $= {[x^{2} l n (16)]}_{4}^{3 + \sqrt{30}} - {[\frac{x^{2}}{2} l n (x) - \frac{x^{2}}{4}]}_{4}^{3 + \sqrt{30}}$
	Commented by cesar.marval.larez@gmail.com last updated on 31/May/19

	${H a v e n}^{'} t g r a p h ?$
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