Question

Calculate the area bounded by the curves `y=x^2-4x+5` and `y=-2x+8`?

Answer 1

`32/3`

Explanation:

graph{(y-(x^2-4x+5))(y-(-2x+8))=0 [-5, 5, -5, 12]}

First we solve the simultaneous equations

`{ (y=x^2-4x+5), (y=-2x+8) :}`

Which requires that:

`x^2-4x+5 = -2x+8`
`:. x^2 -2x-3 = 0`
`:. (x+1)(x-3)= 0`
`:. x=-1,=3`

Hence, the area sought is given by:

`A = int_(-1)^(3) \ (-2x+8) - (x^2-4x+5) \ dx`
`\ \ = int_(-1)^(3) \ -2x+8 - x^2+4x-5 \ dx`
`\ \ = int_(-1)^(3) \ - x^2+2x+3 \ dx`
`\ \ = [ - x^3/3 + x^2 +3x ]_(-1)^(3)`
`\ \ = (-9 +9+9)-(1/3+1-3)`
`\ \ = 9-(-5/3)`
`\ \ = 32/3`