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

To integrate this, we must employ Integration by Parts.

"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).

Then plugging into the IBP formula gives us:

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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 ).

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