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Math Help - Is it just me, or evaluating integrals is not the easist thing to do? Please help!

  1. #1
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    Unhappy [SOLVED] Is it just me, or evaluating integrals is not the easist thing to do?

    This is my first time evaluating integrals on my own and, unfortunately, I just can't seem to get it. As easy as they may seem, integrals are giving me such a hard time, even the simplest ones are proving to be a challenge.

    I need help evaluating the integral below:



    I ended up with 15ln(-1)-15ln(-3)=DNE, but I'm almost certain I got it wrong.
    Last edited by theunforgiven; December 4th 2012 at 04:37 AM.
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    Re: Is it just me, or evaluating integrals is not the easist thing to do? Please help

    You should have a list of standard integrals, if you haven't got one then get one. Later, it helps if you can commit them to memory.
    The one you need here is \int \frac{1}{x}\, dx = \log_{e}x + C.
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    Re: Is it just me, or evaluating integrals is not the easist thing to do? Please help

    Actually it's \displaystyle \begin{align*} \int{\frac{1}{x}\,dx} = \ln{|x|} + C \end{align*}
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    Re: Is it just me, or evaluating integrals is not the easist thing to do? Please help

    Quote Originally Posted by Prove It View Post
    Actually it's \displaystyle \begin{align*} \int{\frac{1}{x}\,dx} = \ln{|x|} + C \end{align*}
    That's what I was about to say. Where do I take it from there though? I know that one would be 15ln{|x|} + C but the numbers, -1 and -3, on that symbol are confusing the hell out of me!
    If I knew what to do, I wouldn't be asking...
    Last edited by theunforgiven; December 4th 2012 at 03:46 AM.
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    Re: Is it just me, or evaluating integrals is not the easist thing to do? Please help

    The absolute value turns anything negative positive.
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    Re: Is it just me, or evaluating integrals is not the easist thing to do? Please help

    Quote Originally Posted by Prove It View Post
    The absolute value turns anything negative positive.
    Oh hold on a sec! Would the answer be:
    15ln|-1|-15ln|-3|=-16 (approximated)
    ?
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    Re: Is it just me, or evaluating integrals is not the easist thing to do? Please help

    Quote Originally Posted by Prove It View Post
    Actually it's \displaystyle \begin{align*} \int{\frac{1}{x}\,dx} = \ln{|x|} + C \end{align*}
    Agreed, lazy of me.
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    Re: Is it just me, or evaluating integrals is not the easist thing to do? Please help

    Quote Originally Posted by theunforgiven View Post
    Oh hold on a sec! Would the answer be:
    15ln|-1|-15ln|-3|=-16 (approximated)
    ?
    No. Like I said, absolute values turn anything negative positive. So \displaystyle \begin{align*} 15\ln{|-1|} - 15\ln{|-3|} = 15\ln{(1)} - 15\ln{(3)} = 0 - 15\ln{(3)} = -15\ln{(3)} \end{align*}
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    Re: Is it just me, or evaluating integrals is not the easist thing to do? Please help

    Quote Originally Posted by Prove It View Post
    No. Like I said, absolute values turn anything negative positive. So \displaystyle \begin{align*} 15\ln{|-1|} - 15\ln{|-3|} = 15\ln{(1)} - 15\ln{(3)} = 0 - 15\ln{(3)} = -15\ln{(3)} \end{align*}
    And that's exactly what I did. I just didn't show that extra step where you put the positive values between parentheses. I did, however, turn them to positive and put them between parentheses when I plugged in the equation into my calculator. But yeah, thank you so much!
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    Re: Is it just me, or evaluating integrals is not the easist thing to do? Please help

    Quote Originally Posted by theunforgiven View Post
    Oh hold on a sec! Would the answer be:
    15ln|-1|-15ln|-3|=-16 (approximated)
    ?
    Yes, although -16.5 would be a better approximation.

    One of the first things you should have learned about "definite integrals" (actually, the definition) is that if F is an anti-derivative of f (that is, if dF/dx= f) then \int_a^b f(x)dx= F(b)- F(a).

    Actually, yes, finding anti-derivatives is, typically harder than differentiating- although evaluating definite integrals should not be more difficult than evaluating any functions. There is a general distinction in mathematics between "direct problems", where we are given a specific definition or formula, and "inverse problems" where we are to "reverse" some direct problem. For example, if I told you that y= x^5- 3x^4+ 4x^2- 7x+ 9 and ask you to find y when x= 3, that would be easy- just do the arithmetic in that formula. But if I told you that y= 3 and asked you to find x,, that would be a much more difficult problem- and there might be many solutions or none.

    We learn, in introductory Calculus, a formula for the derivative of a function so finding a derivative can always go back to using that formula (though we can also use the many "theorems" for special cases derived from that formula). That's a direct problem. But an anti-derivative of f, say, is only defined as "a function whose derivative is f". We don't have a direct formula to use, we have to try to "remember" a function whose derivative is the given function. That's the "inverse" problem.
    Last edited by HallsofIvy; December 4th 2012 at 06:14 AM.
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    Re: Is it just me, or evaluating integrals is not the easist thing to do? Please help

    Quote Originally Posted by HallsofIvy View Post
    Yes, although -16.5 would be a better approximation.

    One of the first things you should have learned about "definite integrals" (actually, the definition) is that if F is an anti-derivative of f (that is, if dF/dx= f) then \int_a^b f(x)dx= F(b)- F(a).

    Actually, yes, finding anti-derivatives is, typically harder than differentiating- although evaluating definite integrals should not be more difficult than evaluating any functions. There is a general distinction in mathematics between "direct problems", where we are given a specific definition or formula, and "inverse problems" where we are to "reverse" some direct problem. For example, if I told you that y= x^5- 3x^4+ 4x^2- 7x+ 9 and ask you to find y when x= 3, that would be easy- just do the arithmetic in that formula. But if I told you that y= 3 and asked you to find x,, that would be a much more difficult problem- and there might be many solutions or none.

    We learn, in introductory Calculus, a formula for the derivative of a function so finding a derivative can always go back to using that formula (though we can also use the many "theorems" for special cases derived from that formula). That's a direct problem. But an anti-derivative of f, say, is only defined as "a function whose derivative is f". We don't have a direct formula to use, we have to try to "remember" a function whose derivative is the given function. That's the "inverse" problem.
    That helped immensely! Thank you
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