Integral Calculus... 4 ò [g(x)] g'(x) dx =?

3 ⤊

2006-10-26 15:50:14 · 4 answers · asked by Olivia 4 in Science & Mathematics ➔ Mathematics

I got... 2[g(x)]2g'(x) + C

2006-10-26 15:50:41 · update #1

Let u = g(x). Then du = g'(x) dx, so you have

4 integral of u du = 4u^2/2 + C = 2u^2 + C = 2[g(x)]^2 + C.

2006-10-26 15:55:55 · answer #1 · answered by James L 5 · 1⤊ 1⤋

u=g(x)
du= g'(x) dx

integral of [g(x)] g'(x) dx = integral of udu = u^2 /2 + C
= g(x)^2 /2 +C

2006-10-27 03:30:53 · answer #2 · answered by locuaz 7 · 0⤊ 1⤋

No. Let u=g(x), so du=g'(x)dx. You will get 2[g(x)]^2 +C

2006-10-26 15:55:07 · answer #3 · answered by mathematician 7 · 2⤊ 2⤋

4â«g(x) . g'(x) dx
= 2 [g(x)]Â² + c

2006-10-26 15:55:52 · answer #4 · answered by Wal C 6 · 0⤊ 2⤋