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

How do you evaluate the integral int 1/x^3dxint 1/x^3dx from 33 to oooo?