Find the definate integal (upper limit 1, lower limit 0) (2x^2-4x-6)dx

Answer 1

what is limit 2(x-3)(x+1)dx

Isn't this a integral uv dv problem?

Answer 2

Looks like a straight polynomial integral to me.

Answer 3

Integral (2x^2-4x-6)dx from 0 to 1
= (2/3)x^3 - 2x^2 - 6x from 0 to 1
= (2/3)(1)^3 - 2(1)^2 - 6(1) - ((2/3)(0)^3 - 2(0)^2 - 6(0))
= 2/3 - 2 - 6 - (0 - 0 - 0)
= 2/3 - 8
= 2/3 - 24/3
= -22/3

Hope this helps :3