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