Question

How do I evaluate this integral.

integrate from -2 to 0: (u^5-u^3+u^2)

Answer 1

I get the indefinite integral u^6/6-u^4/4+U^3/3

Answer 2

use the standard algebraic integral

that correct

Answer 3

Yea, but when I evaluate it I am getting 0-[~-11+4~-3] = ~10 answer is -4 though...

Answer 4

its 0 -[ (-2)^6/6 - (-2)^4/4 + (-2)^3 /3]
= 0 -(64/6 - 16/4 -8/3]

Answer 5

your integration is right...so look again at the arithmetic.\[0-[(1/6)(64) - (1/4)(16) +(1/3)(-8)]\]\[-[(32/3)-(12/3)-(8/3)]\]\[-(12/3)=-4\]

Answer 6

yeah = -4

Answer 7

1/6u^6-1/4u^4+1/3u^3--(-2,0)=-4

Answer 8

Oh I see I lost the negative for +-12/3