Find the area bounded by the curve #y=2x^2-6x# and #y=-x^2+9#?

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Answer 1 · 2018-03-15T10:37:10

#"Area" = 32#

First, it is a good idea to graph the given curves:

Desmos

We need to evaluate the area bounded by these curves, i.e find the area in between them.

That's the area between their points of intersection, so we need to find that: their points of intersection:

#Rightarrow 2x^(2) - 6x = - x^(2) + 9#

#Rightarrow 3x^(2) - 6x - 9 = 0#

#Rightarrow x^(2) - 2x - 3 = 0#

#Rightarrow x^(2) + x - 3x - 3 = 0#

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

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

#therefore x = - 1, 3#

We don't really need to find the points of intersection; the #x#-intercepts will suffice.

This is because the curves are already specified in terms of #y#.

Now we can start evaluating the definite integrals of these curves in the interval #[- 1, 3]#:

#Rightarrow int_(- 1)^(3)# #((- x^(2) + 9) - (2x^(2) - 6x))# #dx#

#= int_(- 1)^(3)# #(- 3x^(2) + 6x + 9)# #dx#

#= |- x^(3) + 3x^(2) + 9x|_(- 1)^(3)#

#= (- (3)^(3) + 3(3)^(2) + 9(3)) - (- (- 1)^(3) + 3(- 1)^(2) + 9(- 1))#

#= (- 27 + 27 + 27) - (1 + 3 - 9)#

#= (27) - (- 5)#

#= 32#

Therefore, the area bounded by the curves is #32#.