Question #0993b

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Answer 1 · 2017-05-05T11:31:05

#16/3#sq. units

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Given: #y^2=2x+3# and #y = x#

Here is a graph of the two equations:

graph{(2x+3-y^2)(y-x)=0 [-10, 10, -5, 5]}

Solving these equations:

#y^2=2x+3" [1]"#
#y = x" [2]"#

Substituting equation [2] into equation [1]:

#x^2=2x+3#

#x^2 - 2x - 3=0#

#(x-3)(x+1)=0#

#x = 3 and x = -1#

#:. "Area" = int_-1^3(y - (y^2-3)/2)dy#

#= {:y^2/2-y^3/6+3/2y]_-1^3#

#= 16/3 "units"^2#