`y_1 = x^2 -4x + 3 , y_2 = -x^2 + 2x + 3` Set up the definite integral that gives the area of the region

Given the curve equations ,they are

`y_1 = x^2 -4x + 3 ` -----(1)

` y_2 = -x^2 + 2x + 3` -----(2)

to get the boundaries or the intersecting points of the curves we have to equate the curves .

`y_1=y_2`

=> `x^2 -4x + 3 = -x^2 + 2x + 3`

=> `2x^2-6x =0`

=> `2x(x-3)=0`

=> `x=0 or x=3`

Area = `int_0^3 ((-x^2 + 2x + 3)-(x^2 -4x + 3)) dx `

= `int _0 ^3 (6x -2x^2) dx `

= `[6x^2 /2 -2x^3 /3]_0 ^3`

= `[6(3)^2 /2 -2(3)^3 /3]-[0]`

=`3^3 -2*3^2 = 27-18 = 9`

so the area of the region between the curves is `9`

Comments

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