What is the area cut off the parabola 4y = 3𝑥^2 by the line 2y = 3x + 12 ?

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Answer 1 · 2018-07-28T13:14:31

#44.859\ \text{unit}^2#

Setting #y=3/4x^2# in the equation of straight line: #2y=3x+12#

#2(3/4x^2)=3x+12#

#x^2-2x-8=0#

#(x-4)(x+2)=0#

#x=4, -2#

The corresponding values of #y# are

#y=12, 3#

hence the line: #2y=3x+12# intersects the parabola at two points #(4, 12)# & #(-2, 3)#

Now, area region enclosed by the parabola & line is given as

#\int (y_1-y_2)\ dx#

Setting #y_1=3/2x+6# & #y_2=3/4x^2# & using proper limits from #x=-2# to #x=4#,

#=\int_{-2}^4 (3/2x+6-3/4x^2)\ dx#

#=(3/4x^2+6x-1/x^3)_{-2}^4#

#=3/4(4^2-(-2)^2)+6(4-(-2))-(1/(4)^3-1/{(-2)^3})#

#=2871/64#

#=44.859\ \text{unit}^2#