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  • Find the area enclosed by the parabola 4y = 3x^2 and the line 2y = 3x + 12.

Q : 7 Find the area enclosed by the parabola \small 4y=3x^2 and the line \small 2y=3x+12.

Answers (1)

best_answer

We have to find the area of the shaded region COB

\\2y=3x+12\\ \Rightarrow y=\frac{3}{2}x+6\\ 4y=3x^{2}\\ \Rightarrow y=\frac{3x^{2}}{4}

The two curves intersect at points (2,3) and (4,12)

Required area is 

\\\int_{-2}^{4}(\frac{3}{2}x+6-\frac{3x^{2}}{4})dx\\ =[\frac{3x^{2}}{4}+6x-\frac{x^{3}}{4}]{_{-2}}^{4}\\ =[12+24-16]-[3-12+2]\\ =20-(-7)\\ =27\ units

Posted by

Sayak

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