elongated S (sum) (x + 3)(x - 1)^10 dx =
2006-10-26 15:41:53 · 4 answers · asked by Olivia 4 in Science & Mathematics ➔ Mathematics
or
ò (x + 3)(x - 1)10 dx =
2006-10-26 15:43:23 · update #1
Let u=x-1. Then du=dx, and you have
integral of (u+4)u^10 du =
integral of u^11 + 4u^10 du =
u^12 / 12 + 4u^11/11 + C =
(x-1)^12 / 12 + 4(x-1)^11 / 11 + C
2006-10-26 15:44:43 · answer #1 · answered by James L 5 · 1⤊ 0⤋
haha
elongated 's'
i find that humorous for some reason
oook
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2006-10-26 22:46:12 · answer #2 · answered by Anonymous · 0⤊ 1⤋
let u=(x+3) dv=(x-1)^10 now the integral
=(x+3)(x-1)^11/11-integral (x-1)^11/11dx
=(1/11)(x+3)(x-1)^11-(1/11)(x-1)^12/12
(1/1)1(x+3)(x-1)^11-1/132(x-1)^12+C
2006-10-26 22:46:42 · answer #3 · answered by raj 7 · 1⤊ 0⤋
(x+3)(x-1)^11
------------------- dx^2
11
2006-10-26 22:57:41 · answer #4 · answered by diskedit 1 · 0⤊ 0⤋