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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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