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 .


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Question Number 73715 by Learner-123 last updated on 15/Nov/19

$E v a l u a t e t h e i n t e g r a l :$ $\int_{R} \int (3 x^{2} + 14 x y + 8 y^{2}) d x d y f o r t h e r e g i o n$ $R in t h e 1 s t q u a d r a n t b o u n d e d b y t h e$ $l i n e s y = \frac{- 3}{2} x + 1, y = \frac{- 3}{2} x + 3, y = - \frac{1}{4} x$ $a n d y = - \frac{1}{4} x + 1 .$
Commented by Learner-123 last updated on 15/Nov/19

$I s t h e a n s . (11.02) ?$
Commented by Joel578 last updated on 15/Nov/19

$If the region only in 1^{st} quadrant, then$ $we {won}^{'} t use y = - \frac{1}{4} x$
	Answered by Joel578 last updated on 15/Nov/19

	Commented by Joel578 last updated on 15/Nov/19

	$I = \underset{R}{\int \int} (3 x^{2} + 14 x y + 8 y^{2}) d y d x$ $= \int_{0}^{\frac{2}{3}} \int_{\frac{- 3}{2} x + 1}^{\frac{- 1}{4} x + 1} (3 x^{2} + 14 x y + 8 y^{2}) d y d x$ $+ \int_{\frac{2}{3}}^{\frac{8}{5}} \int_{0}^{\frac{- 1}{4} x + 1} (3 x^{2} + 14 x y + 8 y^{2}) d y d x$ $+ \int_{\frac{8}{5}}^{2} \int_{0}^{\frac{- 3}{2} x + 3} (3 x^{2} + 14 x y + 8 y^{2}) d y d x$
	Commented by Learner-123 last updated on 16/Nov/19

	$t h a n k s s i r .$
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