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Question Number 162054 by amin96 last updated on 25/Dec/21

	Answered by aleks041103 last updated on 25/Dec/21
	$∫∫_(R) ydxdy = I R={(x,y)∣2x<y<3−x^2 , −3<x<1} ⇒I=∫∫_(R) ydxdy=∫_(−3) ^( 1) (∫_(2x) ^( 3−x^2 ) ydy)dx ∫_(2x) ^( 3−x^2 ) ydy=(1/2)y^2 ∣_(2x) ^(3−x^2 ) =(1/2)((3−x^2 )^2 −4x^2 )= =(1/2)(9+x^4 −6x^2 −4x^2 )=(1/2)x^4 −5x^2 +(9/2) ⇒I=∫_(−3) ^( 1) ((1/2)x^4 −5x^2 +(9/2))dx= =((1/(10))x^5 −(5/3)x^3 +(9/2)x)_(−3) ^1 = =((244)/(10))−((140)/3)+18= =((122.3−140.5)/(15))+18= =((366−700)/(15))+18= =((18.15−334)/(15))=((270−334)/(15))=−((64)/(15))$
	$\underset{R}{\int \int} y d x d y = I$ $R = {(x, y) ∣ 2 x < y < 3 - x^{2}, - 3 < x < 1}$ $\Rightarrow I = \underset{R}{\int \int} y d x d y = \int_{- 3}^{1} (\int_{2 x}^{3 - x^{2}} y d y) d x$ $\int_{2 x}^{3 - x^{2}} y d y = \frac{1}{2} y^{2} ∣_{2 x}^{3 - x^{2}} = \frac{1}{2} ({(3 - x^{2})}^{2} - 4 x^{2}) =$ $= \frac{1}{2} (9 + x^{4} - 6 x^{2} - 4 x^{2}) = \frac{1}{2} x^{4} - 5 x^{2} + \frac{9}{2}$ $\Rightarrow I = \int_{- 3}^{1} (\frac{1}{2} x^{4} - 5 x^{2} + \frac{9}{2}) d x =$ $= {(\frac{1}{10} x^{5} - \frac{5}{3} x^{3} + \frac{9}{2} x)}_{- 3}^{1} =$ $= \frac{244}{10} - \frac{140}{3} + 18 =$ $= \frac{122.3 - 140.5}{15} + 18 =$ $= \frac{366 - 700}{15} + 18 =$ $= \frac{18.15 - 334}{15} = \frac{270 - 334}{15} = - \frac{64}{15}$
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