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Math Help - New to Integrating, I need a lot of help

  1. #16
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by R3ap3r View Post
    uh my answer i got is 6.472. Am I even in the ballpark?
    nope. you're outside, in the parking lot. freezing your tush off.

    Quote Originally Posted by Babs0201 View Post
    I got 32.9983 as an answer. Hope that helps.
    this is correct
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  2. #17
    Junior Member R3ap3r's Avatar
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    Where do I go after I have


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  3. #18
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    Quote Originally Posted by R3ap3r View Post
    Where do I go after I have

    I would just put it in a calculator
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  4. #19
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    Quote Originally Posted by R3ap3r View Post
    Where do I go after I have

    What you should have is not -1 and 5, but u(-1) and u(5).
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  5. #20
    Junior Member R3ap3r's Avatar
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    Quote Originally Posted by icemanfan View Post
    What you should have is not -1 and 5, but u(-1) and u(5).
    What does u(5) signify?
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  6. #21
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by R3ap3r View Post
    What does u(5) signify?
    u is a function of x. u(5) means you plug 5 into the function and evaluate it.

    several substitutions were offered for this problem. which did you use?
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  7. #22
    Junior Member R3ap3r's Avatar
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    Hmm I c what i did wrong atleast. How exactly do you evaluate it?
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    Quote Originally Posted by R3ap3r View Post
    Hmm I c what i did wrong atleast. How exactly do you evaluate it?
    It depends on what your substitution is. for instance, if you used the substitution u = \sqrt{x^3 + 3}, then you'd just plug in

    when x = 5, then u = \sqrt{5^3 + 3} = \sqrt{128} = u(5)

    when x = -1, then u = \sqrt{(-1)^3 + 3} = \sqrt{2} = u(-1)
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  9. #24
    Junior Member R3ap3r's Avatar
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    ohhhh let me try that out on my calculator

    hmmm i got 2\cdot(\sqrt{128})^{\frac{1}{2}} - 2\cdot(\sqrt{2})^{\frac{1}{2}} which led me to 4.348. What am I doing wrong?
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  10. #25
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by R3ap3r View Post
    ohhhh let me try that out on my calculator

    hmmm i got 2\cdot(\sqrt{128})^{\frac{1}{2}} - 2\cdot(\sqrt{2})^{\frac{1}{2}} which led me to 4.348. What am I doing wrong?
    i have no idea.

    the integral you are using is \frac {10}3 \int 1~du, I assume

    then you should get \frac {10u}3 \bigg|_{\sqrt{2}}^{\sqrt{128}} = \frac {10 \sqrt{128}}3 - \frac {10 \sqrt{2}}3 = 32.9983

    maybe you should start from the beginning. i think you're mixing yourself up in the middle somewhere. what substitution are you using?
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  11. #26
    Junior Member R3ap3r's Avatar
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    well I was trying to apply what icemanfan said earlier about evaluating so I plugged the numbers into what he gave me which I suppose I shouldn't do. Now I'm just really confused.
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  12. #27
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    Quote Originally Posted by R3ap3r View Post
    well I was trying to apply what icemanfan said earlier about evaluating so I plugged the numbers into what he gave me which I suppose I shouldn't do. Now I'm just really confused.
    Ok. let's start over. Pick a substitution that you like and we will work through it
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    Junior Member R3ap3r's Avatar
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  14. #29
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    Hello,

    Ok, now calculate \frac{du}{dx}

    The derivative for such a thing is \frac{u'(x)}{2 \sqrt{u(x)}}

    Thus \frac{du}{dx}=\frac{3x^2}{2\sqrt{x^3+3}} \Longrightarrow dx=du \frac{2\sqrt{x^3+3}}{3x^2}=du \frac{2 u}{3x^2}

    Now replacing in the integral :

    \int_{-1} ^5\;\frac{5x^2}{\sqrt{x^3+3}}\;dx=\int_{-1}^5 \frac{5x^2}{u} dx

    Replacing dx :

    \int_{-1}^5 \frac{5{\color{red}x^2}}{\color{blue}u} \frac{2 {\color{blue}u}}{3{\color{red}x^2}} du= \dots
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  15. #30
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by R3ap3r View Post
    Okay. so let's get the limits out of the way.

    we already saw that the new limits for u are \sqrt{128} and \sqrt{2}. so good.

    Now, u = \sqrt{x^3 + 3}

    \Rightarrow du = \frac {3x^2}{2 \sqrt{x^3 + 3}}~dx .............(by the chain rule)

    \Rightarrow \frac 23 ~du = \frac {x^2}{\sqrt{x^3 + 3}}~dx

    \Rightarrow \frac {10}3~du = \frac {5x^2}{\sqrt{x^3 + 3}}~dx ..........................multiplied both sides by 5

    What Luck! the right hand side is exactly our integral. this does not always happen, but it is really nice here, and we won't complain. so we can simply replace our entire integral with 10/3 in du.

    So, our integral becomes:

    \int_{\sqrt{2}}^{\sqrt{128}} \frac {10}3~du = \frac {10}3u \bigg|_{\sqrt{2}}^{\sqrt{128}} = \frac {10}3 (\sqrt{128} - \sqrt{2})

    and then just plug that into your calculator

    do you get it?
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