2007-01-13 14:56:08 · 7 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Integration (7+3x^3/2)/root of x *dx
= Integration 7/root of x *dx + Integration (3x^3/2)/root of x *dx
= 14 * root of x + Integration 3x *dx
= 14 * root of x + 3/2 x^2
Did I understand correctly the question?
2007-01-13 15:06:52 · answer #1 · answered by roman_king1 4 · 0⤊ 1⤋
/root of x = x^-1/2 so integrate 7x^-1/2 + 3x. Do each term separately so Integral = 14 x^1/2 + 3/2x^2
2007-01-13 15:03:50 · answer #2 · answered by hello 6 · 0⤊ 1⤋
a) use substitution (you may exchange the obstacles of integration in accordance to the contemporary function in terms of u) u = 3x^2 du = 6x a million/6du = dx rewrite in terms of u and are available to a determination c) discover the element the situation (x-2) alterations from unfavourable to useful (at x = 2) combine an identical function (with out abs value) from 0 to 2, and multiply the effect by utilising utilising -a million. Then combine 2 to 3 regularly. upload those 2 outcomes
2016-12-16 04:08:52 · answer #3 · answered by binford 4 · 0⤊ 0⤋
Simplify it first:
Integral(7/rootX)dx + Integral(3X)dx
Much easier now huh???
This becomes: 14rootX + 3/2 * X^2
2007-01-13 15:32:56 · answer #4 · answered by Stryder 2 · 0⤊ 0⤋
(7+3x^3/2)/root of x *dx
=7x ^ -1/2 + 3x^1/2 *dx
integral[7x ^ -1/2 + 3x^1/2 *dx]
=14x^1/2 + 2x^3/2 + C
2007-01-13 15:03:49 · answer #5 · answered by Anonymous · 0⤊ 1⤋
integrale of(7/Vx)dx+integrale of x^3/2)/Vxdx= integrale of7/x^1/2dx+integrale of3x^3/2/x^1/2=
'' =-7/2 root of x+3x +k
2007-01-13 15:14:50 · answer #6 · answered by Johnny 2 · 0⤊ 0⤋
I use my Ti -89 to solve that :)
2007-01-13 15:02:58 · answer #7 · answered by NFLS121a 1 · 0⤊ 1⤋