# Thread: limit as lower bound of an integral goes to infinity

1. ## limit as lower bound of an integral goes to infinity

We are given the integral
$\displaystyle \int_a^b x^2\,dx =2$
We are then asked to explain, intuitively, what happens to the length of the interval (a,b) as a approaches infinity.

Intuitively, the interval is getting shorter and shorter as a goes to infinity. Easy enough, but here's where I'm stuck. We then are asked to mathematically justify this by finding

$\displaystyle \displaystyle \lim_{a\to\infty}(b-a)$

We are given a hint that is to write either a or b in terms of the other, then write the expression in terms of one variable before taking the limit. Sounds easy, but this is what I've got so far.

$\displaystyle \displaystyle \int_a^b x^2\,dx = 2\quad \Rightarrow\quad\frac{b^3}{3} - \frac{a^3}{3} = 2\quad\Rightarrow\quad b^3 - a^3 = 6\quad\Rightarrow\quad b = \sqrt[3]{a^3 + 6}$

Plugging this back in and evaluating the limit is where I get stuck.

$\displaystyle \displaystyle \lim_{a\to\infty}(b-a) = \lim_{a\to\infty}\left(\sqrt[3]{a^3+6} - a^3\right)$

I'm not really sure how to evaluate this limit.

^^^^^^^^^^^^^^^^^^^^^^^^^

Is there something to be done here by stating that

$\displaystyle a^3 > \sqrt[3]{a^3+6}$

I don't know where to go with this. Is there some fancy completing the cube action that has to take place to get that a^3 out of the radical?

Even doing
$\displaystyle \displaystyle \sqrt[3]{a^3 + 6} - a^3 = a\sqrt[3]{1-\frac{6}{a^3}} - a^3$

Am I at least on the right track? Or am I completely setting this up wrong?

Who woulda thought I'd be asked to mathematically justify something in a math class. Doh!

2. Originally Posted by pirateboy
We are given the integral
$\displaystyle \int_a^b x^2\,dx =2$
We are then asked to explain, intuitively, what happens to the length of the interval (a,b) as a approaches infinity.

Intuitively, the interval is getting shorter and shorter as a goes to infinity. Easy enough, but here's where I'm stuck. We then are asked to mathematically justify this by finding

$\displaystyle \displaystyle \lim_{a\to\infty}(b-a)$

We are given a hint that is to write either a or b in terms of the other, then write the expression in terms of one variable before taking the limit. Sounds easy, but this is what I've got so far.

$\displaystyle \displaystyle \int_a^b x^2\,dx = 2\quad \Rightarrow\quad\frac{b^3}{3} - \frac{a^3}{3} = 2\quad\Rightarrow\quad b^3 - a^3 = 6\quad\Rightarrow\quad b = \sqrt[3]{a^3 + 6}$

Plugging this back in and evaluating the limit is where I get stuck.

$\displaystyle \displaystyle \lim_{a\to\infty}(b-a) = \lim_{a\to\infty}\left(\sqrt[3]{a^3+6} - a^3\right)$

You're subtracting "a^3" rather than "a".

I'm not really sure how to evaluate this limit.

^^^^^^^^^^^^^^^^^^^^^^^^^

Is there something to be done here by stating that

$\displaystyle a^3 > \sqrt[3]{a^3+6}$

I don't know where to go with this. Is there some fancy completing the cube action that has to take place to get that a^3 out of the radical?

Even doing
$\displaystyle \displaystyle \sqrt[3]{a^3 + 6} - a^3 = a\sqrt[3]{1-\frac{6}{a^3}} - a^3$

Shouldn't you be subtracting "a" ?, not it's cube.

Am I at least on the right track? Or am I completely setting this up wrong?

Who woulda thought I'd be asked to mathematically justify something in a math class. Doh!
.

3. Err...unless I can do this?

$\displaystyle \displaystyle \lim_{a\to\infty}\left(\sqrt[3]{a^3 + 6} - a^3\right) = \lim_{a\to\infty}\left(a\sqrt[3]{1-\frac{6}{a^3}} - a^3\right) = \lim_{a\to\infty}\left[a^3\left(\frac{1}{a^2}\sqrt[3]{a^3 + 6} - 1\right)\right]$
$\displaystyle \displaystyle = \lim_{a\to\infty}(a^3) \left(\lim_{a\to\infty}\frac{1}{a^2}\sqrt[3]{a^3 + 6} - 1\right) = \infty(0 - 1) = -\infty$

Haha, I don't think I can do that, though.

4. Originally Posted by pirateboy
Err...unless I can do this?

$\displaystyle \displaystyle \lim_{a\to\infty}\left(\sqrt[3]{a^3 + 6} - a^3\right) = \lim_{a\to\infty}\left(a\sqrt[3]{1-\frac{6}{a^3}} - a^3\right) = \lim_{a\to\infty}\left[a^3\left(\frac{1}{a^2}\sqrt[3]{a^3 + 6} - 1\right)\right]$
$\displaystyle \displaystyle = \lim_{a\to\infty}(a^3) \left(\lim_{a\to\infty}\frac{1}{a^2}\sqrt[3]{a^3 + 6} - 1\right) = \infty(0 - 1) = -\infty$

Haha, I don't think I can do that, though.
$\displaystyle \displaystyle \lim_{a\to\infty}\left(\sqrt[3]{a^3 + 6} - a\right) = \lim_{a\to\infty}\left(a\sqrt[3]{1-\frac{6}{a^3}} - a\right) =a-a$

5. Ah yeah...I should've been considering 'a', not 'a^3'. Cool thanks! So what this is saying is that as 'a' goes to infinity the width of the interval becomes narrower and narrower and eventually becomes a straight line?

6. Originally Posted by pirateboy
Ah yeah...I should've been considering 'a', not 'a^3'. Cool thanks! So what this is saying is that as 'a' goes to infinity the width of the interval becomes narrower and narrower and eventually becomes a straight line?
That's how it certainly ends up looking!