K kangarulz Procrastinaton Joined Mar 23, 2005 Messages 144 Gender Female HSC 2006 Jul 13, 2006 #1 integrate with respect to x: xe^(x^2) + xe^(-x^2)
R Riviet . Joined Oct 11, 2005 Messages 5,584 Gender Undisclosed HSC N/A Jul 13, 2006 #2 I'll start you off: Let u=x2 du/dx = 2x x.ex2 + x.e-x2 = 1/2.(2x.ex2) + 1/2.(2x.ex2) Hope that helps. Last edited: Jul 13, 2006
I'll start you off: Let u=x2 du/dx = 2x x.ex2 + x.e-x2 = 1/2.(2x.ex2) + 1/2.(2x.ex2) Hope that helps.
hyparzero BOS Male Prostitute Joined Sep 10, 2005 Messages 241 Location Wankersville Gender Female HSC 2006 Jul 13, 2006 #3 Riviet said: I'll start you off: Let u=x2 du/dx = 2x x.ex2 + x.e-x2 = 1/2.(2x.ex2) + 1/2.(2x.ex2) Hope that helps. Click to expand... no it doesnt
Riviet said: I'll start you off: Let u=x2 du/dx = 2x x.ex2 + x.e-x2 = 1/2.(2x.ex2) + 1/2.(2x.ex2) Hope that helps. Click to expand... no it doesnt
alcalder Just ask for help Joined Jun 26, 2006 Messages 597 Location Sydney Gender Female HSC N/A Jul 13, 2006 #4 kangarulz said: integrate with respect to x: xe^(x^2) + xe^(-x^2) Click to expand... OK, the starting point is knowing that: ∫f'(x)ef(x).dx = ef(x) + c SO If we have f(x) = ex2 f'(x) = 2xex2 f(x) = e-x2 f'(x) = -2xe-x2 Now, you can do: ∫ (xex2) + xe-x2)).dx = ∫ (xex2).dx +∫(xe-x2).dx = 1/2∫ (2xex2).dx -1/2∫(2xe-x2).dx = 1/2 (ex2) - 1/2 (e-x2) + c Does that help??
kangarulz said: integrate with respect to x: xe^(x^2) + xe^(-x^2) Click to expand... OK, the starting point is knowing that: ∫f'(x)ef(x).dx = ef(x) + c SO If we have f(x) = ex2 f'(x) = 2xex2 f(x) = e-x2 f'(x) = -2xe-x2 Now, you can do: ∫ (xex2) + xe-x2)).dx = ∫ (xex2).dx +∫(xe-x2).dx = 1/2∫ (2xex2).dx -1/2∫(2xe-x2).dx = 1/2 (ex2) - 1/2 (e-x2) + c Does that help??