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Evaluate-the-integral-R-3x-2-14xy-8y-2-dxdy-for-the-region-R-in-the-1st-quadrant-bounded-by-the-lines-y-3-2-x-1-y-3-2-x-3-y-1-4-x-and-y-1-4-x-1-




Question Number 73715 by Learner-123 last updated on 15/Nov/19
Evaluate the integral :  ∫_( R) ∫(3x^2 +14xy+8y^2 )dxdy for the region  R in the 1st quadrant bounded by the  lines y=((−3)/2)x+1,y=((−3)/2)x+3,y=−(1/4)x  and y=−(1/4)x+1 .
$${Evaluate}\:{the}\:{integral}\:: \\ $$$$\underset{\:\mathbb{R}} {\int}\int\left(\mathrm{3}{x}^{\mathrm{2}} +\mathrm{14}{xy}+\mathrm{8}{y}^{\mathrm{2}} \right){dxdy}\:{for}\:{the}\:{region} \\ $$$$\mathbb{R}\:\mathrm{in}\:{the}\:\mathrm{1}{st}\:{quadrant}\:{bounded}\:{by}\:{the} \\ $$$${lines}\:{y}=\frac{−\mathrm{3}}{\mathrm{2}}{x}+\mathrm{1},{y}=\frac{−\mathrm{3}}{\mathrm{2}}{x}+\mathrm{3},{y}=−\frac{\mathrm{1}}{\mathrm{4}}{x} \\ $$$${and}\:{y}=−\frac{\mathrm{1}}{\mathrm{4}}{x}+\mathrm{1}\:. \\ $$
Commented by Learner-123 last updated on 15/Nov/19
Is the ans.  (11.02) ?
$${Is}\:{the}\:{ans}.\:\:\left(\mathrm{11}.\mathrm{02}\right)\:? \\ $$
Commented by Joel578 last updated on 15/Nov/19
If the region only in 1^(st)  quadrant, then   we won′t use y = −(1/4)x
$$\mathrm{If}\:\mathrm{the}\:\mathrm{region}\:\mathrm{only}\:\mathrm{in}\:\mathrm{1}^{\mathrm{st}} \:\mathrm{quadrant},\:\mathrm{then}\: \\ $$$$\mathrm{we}\:\mathrm{won}'\mathrm{t}\:\mathrm{use}\:{y}\:=\:−\frac{\mathrm{1}}{\mathrm{4}}{x} \\ $$
Answered by Joel578 last updated on 15/Nov/19
Commented by Joel578 last updated on 15/Nov/19
I = ∫∫_(R)  (3x^2  + 14xy + 8y^2 ) dy dx     = ∫_0 ^(2/3) ∫_(((−3)/2)x+1) ^(((−1)/4)x+1)  (3x^2  + 14xy + 8y^2 ) dy dx                 + ∫_(2/3) ^(8/5) ∫_0 ^(((−1)/4)x+1)  (3x^2  + 14xy + 8y^2 ) dy dx                              + ∫_(8/5) ^( 2) ∫_0 ^(((−3)/2)x+3)  (3x^2  + 14xy + 8y^2 ) dy dx
$${I}\:=\:\underset{{R}} {\int\int}\:\left(\mathrm{3}{x}^{\mathrm{2}} \:+\:\mathrm{14}{xy}\:+\:\mathrm{8}{y}^{\mathrm{2}} \right)\:{dy}\:{dx} \\ $$$$\:\:\:=\:\int_{\mathrm{0}} ^{\frac{\mathrm{2}}{\mathrm{3}}} \int_{\frac{−\mathrm{3}}{\mathrm{2}}{x}+\mathrm{1}} ^{\frac{−\mathrm{1}}{\mathrm{4}}{x}+\mathrm{1}} \:\left(\mathrm{3}{x}^{\mathrm{2}} \:+\:\mathrm{14}{xy}\:+\:\mathrm{8}{y}^{\mathrm{2}} \right)\:{dy}\:{dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\:\int_{\frac{\mathrm{2}}{\mathrm{3}}} ^{\frac{\mathrm{8}}{\mathrm{5}}} \int_{\mathrm{0}} ^{\frac{−\mathrm{1}}{\mathrm{4}}{x}+\mathrm{1}} \:\left(\mathrm{3}{x}^{\mathrm{2}} \:+\:\mathrm{14}{xy}\:+\:\mathrm{8}{y}^{\mathrm{2}} \right)\:{dy}\:{dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\:\int_{\frac{\mathrm{8}}{\mathrm{5}}} ^{\:\mathrm{2}} \int_{\mathrm{0}} ^{\frac{−\mathrm{3}}{\mathrm{2}}{x}+\mathrm{3}} \:\left(\mathrm{3}{x}^{\mathrm{2}} \:+\:\mathrm{14}{xy}\:+\:\mathrm{8}{y}^{\mathrm{2}} \right)\:{dy}\:{dx} \\ $$
Commented by Learner-123 last updated on 16/Nov/19
thanks sir.
$${thanks}\:{sir}. \\ $$

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