How do you evaluate the integral #int 1/x^3dx# from #3# to #oo#?

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Answer 1 · 2018-01-24T03:53:10

Using limits to evaluate the improper integral, the answer is #1/18#

This is an improper limit, so we need to use a limit to solve it.

First, let's go ahead and find the antiderivative of #1/x^3# so we can use it later to solve the integral.

#int1/x^3dx = intx^-3dx = (x^-2)/-2 = -1/(2x^2)#

Now, since we can't technically plug in #oo# to an expression like this, we need to use a limit to evaluate the integral, like this:

#int_3^oo1/x^3dx = lim_(b->oo) int_3^b1/x^3dx#

Now, we can evaluate the integral.

# lim_(b->oo) int_3^b1/x^3dx #

#= lim_(b->oo)[-1/(2x^2)]_3^b#

#= lim_(b->oo) (-1/(2b^2)) - (-1/(2(3)^2))#

#= lim_(b->oo) (-1/(2b^2)) + 1/18#

Since the numerator of the first fraction is 1, and the denominator is approaching infinity, the fraction will approach 0. Therefore:

#lim_(b->oo) (-1/(2b^2)) + 1/18#

#= 0 + 1/18#

#= 1/18#

Final Answer