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Math Help - limit as lower bound of an integral goes to infinity

  1. #1
    Junior Member pirateboy's Avatar
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    limit as lower bound of an integral goes to infinity

    We are given the integral
    \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 \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 \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 \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

    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 \sqrt[3]{a^3 + 6} - a^3 = a\sqrt[3]{1-\frac{6}{a^3}} - a^3
    Seems to lead nowhere.

    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!
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  2. #2
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    Quote Originally Posted by pirateboy View Post
    We are given the integral
    \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 \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 \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 \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

    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 \sqrt[3]{a^3 + 6} - a^3 = a\sqrt[3]{1-\frac{6}{a^3}} - a^3
    Seems to lead nowhere.

    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!
    .
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  3. #3
    Junior Member pirateboy's Avatar
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    Err...unless I can do this?

    \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 = \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.
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  4. #4
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    Quote Originally Posted by pirateboy View Post
    Err...unless I can do this?

    \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 = \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 \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
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  5. #5
    Junior Member pirateboy's Avatar
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    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?
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  6. #6
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    Quote Originally Posted by pirateboy View Post
    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!
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