how do you integrate (x^3)(e^-3x)dx
d/dx(x^3e^-3x) = (x^3)(-3e^-3x) + 3x^2(e^-3x)
Is this correct and how do you continue from here?
2 answers
You do it by parts, several times. https://www.integral-calculator.com/ can work it into steps to show you how....it is a busy tablet of work.
let
u = x^n
du = nx^(n-1) dx
dv = e^(ax)
v = 1/a e^(ax)
so,
∫x^n e^(ax) dx = 1/a x^n e^(ax) - n/a ∫x^(n-1) e^(ax) dx
So, to start off,
∫(x^3)(e^-3x)dx = 1/-3 x^3 e^(-3x) - (3/-3)∫x^2 e^(-3x) dx
= -1/3 x^3 e^(-3x) + ∫x^2 e^(-3x) dx
and so on till the x stuff is gone
u = x^n
du = nx^(n-1) dx
dv = e^(ax)
v = 1/a e^(ax)
so,
∫x^n e^(ax) dx = 1/a x^n e^(ax) - n/a ∫x^(n-1) e^(ax) dx
So, to start off,
∫(x^3)(e^-3x)dx = 1/-3 x^3 e^(-3x) - (3/-3)∫x^2 e^(-3x) dx
= -1/3 x^3 e^(-3x) + ∫x^2 e^(-3x) dx
and so on till the x stuff is gone
0 answers