Determine if this integral is convergent. If so, find what value the integral converges to: f(x) = xe-2x^2?

Answer 1

Let u = 2x^2
du = 4x dx

∫ f(x) dx = ∫ 1/4 * e^(-u) du
= -1/4 * e^(-u)

With the limits 0 < u < ∞
Limit = 1/4

Answer 2

Int (0.infinity) x*e^-2x^2 dx
take 2x^2= z so 4x*dx =dz and the integral becomes
Int (0, +infinity) 1/4 e^-z dz = 1/4
If the lower limit of the integral is not 0 but " a "the integral is
1/4*e^(-2a^2)

Answer 3

? xe^(-4x) dx combine by potential of areas dv= e^-4x v=-4e^-4x u=x du=dx ? u dv = u v - ? v dx ? xe^(-4x) dx = -4x e^(-4x) + 4 ? e^(-4x) dx on the cut back x minus infinity e^-4x techniques infinity so -4xe^-4x is diverges we choose no longer analyze the different words because of fact the fundamental has diverged.

Answer 4

let u = -2x^2
du = -4x dx

then you can finish the integral...

its a multiple of e^u


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