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Math Help - Antidifferentiation

  1. #1
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    Antidifferentiation

    Hello, I am having trouble finding the indicated integral on these two problems.

    { (e^x/2 + x times square root of x) dx

    and

    { x^1.1 (1/3x - 1) dx

    For the first one, I had 1(1/2)e^x = 1/2e^x + (1/1+1) (1/(1/2)+1) which doesn't appear to be correct, I also can't figure out the second one. If someone could illustrate how these integrals are found, Thank you.
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  2. #2
    Moo
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    Hello,

    Assuming that i read it correctly...

    \int \frac{e^x}{2}+x \underbrace{\sqrt{x}}_{x^{1/2}} dx

    =\frac{1}{2} \int e^x dx + \int x^{3/2} dx

    =\frac{1}{2} e^x+\frac{2}{5} x^{5/2}
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  3. #3
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    Could you explain this a little more through? I am having trouble mostly with the second term, I don't understand how this turns out to be 2/5 x^5/2 Thanks...
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  4. #4
    Moo
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    Hm,

    The antiderivative for x^n is \frac{x^{n+1}}{n+1}

    Here, how much is n ?

    I'm sorry, this is true i've skipped some steps
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  5. #5
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    Well the way my professor said to do these, was to put the constants in front, (1/2)? and do 1 over p+1 times x^p+1 x^p dx = 1/p+1 X^p+1 So relating this back to this problem, we have x times the square root of x, so the x term would be (1/2)x^2 and the square root is (1/1.5)x^1.5 Is this correct? This still doesn't lead to 2/5x^5/2
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  6. #6
    Moo
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    The x term is x^(3/2) because it's x^1*x^(1/2)=x^(1+1/2)=x^(3/2)

    Here, p=3/2 -> p+1=5/2... But there are details i haven't understood in what you said oO
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  7. #7
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    This is what I am getting for the entire problem:

    1/2e^x + (1/2x^2)(1/(3/2)x^3/2) When I differentiate each one to check my answer, they do in fact differentiate back to the original terms in the problem, so that is what is confusing me, Hope I have explained this well. I just looked closer at your work, and you added the exponents on the x terms, I was multiplying them, but they should be added right?
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  8. #8
    Moo
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    They should be added, yes

    a^b a^c=a^{b+c}

    a^{bc}=(a^b)^c=(a^c)^b

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  9. #9
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    Thanks Moo, it was that small part that had me hung up. How would this problem be intergrated:

    { x^1.1 (1/3x -1) Would I do the same thing as we just did in the previous problem? I started doing it and got (1/2.1)x^2.1 I know the second term we will have the divide by zero error, so this goes to lnx if I am not mistaken? The last term would be (1/2)x^2 but this is all being multiplied...
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  10. #10
    Moo
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    When there is a product, you can't integrate part by part... You can when it's :
    - a multiplication between a constant and a function
    - sum of functions

    Develop it..

    \frac{x^{1.1}}{3x} = \frac{x^{1.1-1}}{3}

    So the integral is \int x^{0.1}-x^{1.1} dx

    And do it just as before
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  11. #11
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    Ok, so if I understand you correctly, it looks like you multiplied x^1.1 by 1/3x, giving us x^1.1 / 3x. The exponent on -1 is 1 so you subtracted it from the exponent 1.1 ? What happened to the denominator of 3 in these terms? I don't understand why x^.1 is subtracting x^1.1 Could you explain this?
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  12. #12
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    Quote Originally Posted by kdogg121 View Post
    Ok, so if I understand you correctly, it looks like you multiplied x^1.1 by 1/3x, giving us x^1.1 / 3x. The exponent on -1 is 1 so you subtracted it from the exponent 1.1 ? What happened to the denominator of 3 in these terms? I don't understand why x^.1 is subtracting x^1.1 Could you explain this?
    \frac{a^b}{a^c}=a^{b-c}
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