Let A be the region bounded by the curves y=x^2-6x+8 and y=0, find the volume when A is revolved around the x-axis

Answer 1

we need the x-intercepts.
x^2 - 6x + 8 - 0
(x-2)(x-4) = 0
x = 2 or x = 4

Vol = π∫y^2 dx from 2 to 4
= π∫(x^2 - 6x + 8)^2 dx
= π∫(x^4 - 12x^3 + 52x^2 - 96x + 64) dx from 2 to 4
= π[x^5/5- 3x^4 + 52x^3/3 - 48x^2 + 64x] from 2 to 4
= π [ (1024/5 - 3(256) + 52(64)/3 - 48(16) + 64(4)0 - (32/5 - 3(160 + 52(8)/3 - 48(4) + 64(2) ) ]
= ...

I will let you finish the buttonpushing.
Also please check my arithmetic, I should have written this down first.