Question

Question #988be

Answer 1

3

Explanation:

Before we start, I'm going to rewrite this expression into something that'll make the expression a little easier to solve:

`int_1^∞x^(-4/3) dx`

This just makes it a bit easier to evaluate the integral later.

Technically, it's not mathematically sound to use infinity as a bound on a definite integral. Hence, what we're gonna do is set the upper bound to be some arbitrary variable (I'm gonna use `b`), and set a limit as that goes to infinity.

So:

`lim_(b->∞)int_1^bx^(-4/3) dx`

Now, we're just going to evaluate this as a definite integral using `b` and `1` as our bounds. The actual integration is a very straightforward antidifferentiation problem, so in the interest of brevity I will skip over that. Once you evaluate the integral, this is what you get:

`lim_(b->∞)-3x^(-1/3) |_1^b`

Applying the Fundamental Theorum of Calculus P2 to evaluate the definite integral:

`=>lim_(b->∞) [(-3b^(-1/3)) - (-3(1)^(-1/3))]`

Now, we need to evaluate the limit. To make this process a bit more clear, I'm gonna rewrite the above expression as follows:

`=>lim_(b->∞) [(-3/b^(1/3)) + 3]`

Now that we have a fraction, evaluating the limit is much, much easier. As the denominator of a fraction goes to infinity, the entire fraction automatically goes to 0. Hence:

`=> cancel((-3/b^(1/3)) + 3`

`=> 3`

There you have it.

If you need some additional help, here are some videos you can watch:

General Overview of Improper Integrals (long, but thorough video)

Antidifferentiation

Fundamental Theorum of Calculus P2

Hope that helped :)