Let f(x)= sqr. rt.(x^2-6x+9) and g(x)=6-f(x). Using integration, what is the number of square units in the area of the region enclosed by the graphs of y=f(x) and y=g(x)?
When f(x)= sqr. rt.(x^2-6x+9) and g(x)=6-f(x), the area enclosed by the graphs of y=f(x) and y=g(x) is 18 square units.
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Let's find the area using integration. The first step is still to observe that
`f ( x ) = sqrt ( x^2 - 6x + 9 ) = sqrt ( ( x - 3 )^2 ) = | x - 3 | .`
It is significant to not forget the absolute value sign.
Now we can write `f ( x ) = {( 3 - x if x lt 3 ) , ( x - 3 if x gt= 3 ):} , `
so `g ( x ) = 6 - f ( x ) = {( 3 + x if x lt 3 ) , ( 9 - x if x gt= 3 ):} .`
The functions are equal where `3 - x = 3 + x , ` that is, at `x_1 = 0 lt 3 ` and where `x - 3 = 9 - x , ` that is, at `x_2 = 6 . ` It is clear that `g ( x ) gt= f ( x ) ` between `x_ 1 ` and `x_2 . ` This way, the area is
`int_(x_1)^(x_2) ( g ( x ) - f ( x ) ) dx = int_0^3 ( g ( x ) - f ( x ) ) dx + int_3^6 ( g ( x ) - f ( x ) ) dx =`
`=int_0^3 2x dx + int_3^6 ( 12 - 2x ) dx = ( x^2 )_(x=0)^3 + ( 12x - x^2 )_(x=3)^6 =`
`= 9 + 12*3 - 36 + 9 = 18 .`
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