Question

Can someone check my work and answer?
Evaluate the integral from -1 to 0 of (4x^6+2x)^3(12x^5+1)dx

My work:
let u=4x^6+2x
dx=du/24x^5+2

now we have the integral from -1 to 0 of u^3(12x^5+1)(du/24x^5+2)
Simplifies to
the integral from -1 to 0 of u^3(1/2+1/2)
simplifies to
the integral from -1 to 0 of u^3

Apply the integral power rule and you'll have
the integral from -1 to 0 of u^4/4
Put the limits into terms of u by pluggin each into the substituted equation. I got 0 for the upper limit and 2 for the lower.

Apply the fundamental theorem of calculus and you'll get 0-4=-4.

Every calculator I've put this in has told me it's -2 though. Can someone show me if this is right or where I went wrong?

Answer 1

OK. Let's see here...
∫(4x^6+2x)^3(12x^5+1)dx
let
u = 4x^6+2x
du = 24x^5+2 = 2(12x^5+1)
so, you now have
∫1/2 u^3 du
= 1/8 u^4 [-1,0]
= 1/8 (4x^6+2x)^4 [-1,0]
= 1/8 (0-(4-2)^4)
= 1/8 (-16)
= -2

where did you get u^3(1/2+1/2) ? There's only the one term.

Another way is to change the limits of integration. Since
u = 4x^6+2x,
when x=-1, u=2
when x=0, u=0
Then we have
∫[-1,0] (4x^6+2x)^3(12x^5+1)dx
= ∫[2,0] 1/2 u^3 du
= 1/8 u^4 [2,0]
= 1/8 (0-16)
= -2

Answer 2

Oh I see. I was dividing 12 by 24 and 1 by 2 to get the 1/2 + 1/2. I didn't even think of factoring out the 2. Thank you!