chosen5 Banned Jan 28, 2003 #1 Integration by Substitution(reverse derivatives) 1: x / (3x+2)^(1/2) 2: 1 / (5x-3)^(1/2) 3: Sin4x 4: (1+x^3)^1/2 * (x^2) 5: (x^2+5)^3
Integration by Substitution(reverse derivatives) 1: x / (3x+2)^(1/2) 2: 1 / (5x-3)^(1/2) 3: Sin4x 4: (1+x^3)^1/2 * (x^2) 5: (x^2+5)^3
Jarconis Board Supporter Jan 28, 2003 #2 you'll never use that in the real world....unless you grow up to be a calc teacher.
Sir Savage Board Sponsor Jan 28, 2003 #3 If I do these problems for you, how does that help you? Besides, "learning" calculus on a forum is not the best way to go about it.
If I do these problems for you, how does that help you? Besides, "learning" calculus on a forum is not the best way to go about it.
YellowJacket Banned Jan 28, 2003 #4 By God, I sucked at calculus, but damnit I can figure your BMR in seconds.
Chemo Brewing Anabolic Minds Jan 28, 2003 #5 Originally posted by chosen5 Integration by Substitution(reverse derivatives) 1: x / (3x+2)^(1/2) 2: 1 / (5x-3)^(1/2) 3: Sin4x 4: (1+x^3)^1/2 * (x^2) 5: (x^2+5)^3 Click to expand... [2*(3x-4) * (3x+2)^.5] / 27 [2*(5x-3)^.5] / 5 -45*cos(4x) / pi [2*(x^3+1)^3/2] / 9 (x^7) / 7 + 3x^5 + 25x^3 + 125x There's your answers (quick and dirty) but the actual work is up to you. Chemo
Originally posted by chosen5 Integration by Substitution(reverse derivatives) 1: x / (3x+2)^(1/2) 2: 1 / (5x-3)^(1/2) 3: Sin4x 4: (1+x^3)^1/2 * (x^2) 5: (x^2+5)^3 Click to expand... [2*(3x-4) * (3x+2)^.5] / 27 [2*(5x-3)^.5] / 5 -45*cos(4x) / pi [2*(x^3+1)^3/2] / 9 (x^7) / 7 + 3x^5 + 25x^3 + 125x There's your answers (quick and dirty) but the actual work is up to you. Chemo