Integration (1 Viewer)

[Saturn WY15]

Prove this by substitution. I don't know why but when i do it it equals 4 but when i do the long method of expanding the whole term it =0

 

[cutemouse]

View attachment 23788

Prove this by substitution. I don't know why but when i do it it equals 4 but when i do the long method of expanding the whole term it =0

Let u=x-2, du/dx=1
When x=4 u=2
When x=0 u=-2

So [maths]I=\int^2_{-2} u^3 \ du = 0[/maths]

(f(u)=u^3, f(-u) = (-u)^3 = -u^3 = -f(u), so f(u) represents an odd function ; The integration of an odd function about symmetrical limits is zero.)