What is the antiderivative of #(x/4)(e^(-x/4))# from 0 to infinity?

Question

William Abdalla · Accepted Answer

# int_0^oo x/4e^(-x/4) dx = 4 #

To integrate this, we must employ Integration by Parts.

# intu(dv)/dxdx = uv - intv(du)/dxdx #, or less formally # intudv=uv-intvdu #

"The integral of udv equals uv minus the integral of vdu" is the less formal rule that I was taught to remember. If you have trouble remembering it, try to visualize it as a direct result of integrating the Product Rule for differentiation.

In essence, we want to find one function that becomes simpler when differentiated and another that becomes simpler when integrated (or at least is integrable).

So for the integrand #x/4e^(-x/4)#, hopefully you can see that #x# simplifies when differentiated and #e^(alphax)# effectively remains unchanged under differentiation or integration.

Let # { (u=x, => , (du)/dx=1), ((dv)/dx=1/4e^(-x/4), =>, v=-e^(-x/4) ) :}#

Then plugging into the IBP formula gives us:

# int(u)((dv)/dx)dx = (u)(v) - int(v)((du)/dx)dx # # :. int (x)(1/4e^(-x/4)) dx = (x)(-e^(-x/4)) - int(-e^(-x/4))(1)dx # # :. int x/4e^(-x/4) dx = -xe^(-x/4) + inte^(-x/4)dx # # :. int x/4e^(-x/4) dx = -xe^(-x/4) -4e^(-x/4) +c # # :. int x/4e^(-x/4) dx = -(x+4)e^(-x/4) +c #

So Applying the limits we have: # \ \ \ \ \ int_0^oo x/4e^(-x/4) dx = -[(x+4)e^(-x/4)]_0^oo # # :. int_0^oo x/4e^(-x/4) dx = -{(lim_(x rarr oo)(x+4)e^(-x/4)) - [(x+4)e^(-x/4)]_0} # # :. int_0^oo x/4e^(-x/4) dx = -(lim_(x rarr oo)(x+4)e^(-x/4)) + 4e^0 # # :. int_0^oo x/4e^(-x/4) dx = -0 + 4 # # :. int_0^oo x/4e^(-x/4) dx = 4 #

William Abdalla · Answer

undefined The antiderivative of $ \frac{x}{4} e^{-\frac{x}{4}} $ with respect to $ x $ is $ -xe^{-\frac{x}{4}} + 4e^{-\frac{x}{4}} + C $, where $ C $ is the constant of integration.

To find the definite integral from 0 to infinity, we evaluate the antiderivative at the upper limit (infinity) and subtract the value of the antiderivative at the lower limit (0):

$$ \lim_{{x \to \infty}} (-xe^{-\frac{x}{4}} + 4e^{-\frac{x}{4}}) - (-0e^{-\frac{0}{4}} + 4e^{-\frac{0}{4}}) $$

$$ = \lim_{{x \to \infty}} (-xe^{-\frac{x}{4}} + 4e^{-\frac{x}{4}}) - 4 $$

Since $ e^{-\frac{x}{4}} $ approaches 0 as $ x $ approaches infinity, we have:

$$ \lim_{{x \to \infty}} (-xe^{-\frac{x}{4}} + 4e^{-\frac{x}{4}}) = 0 $$

So the definite integral from 0 to infinity is simply $ -4 $.