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Thread: Beginner at definite integrals

  1. #1
    s3a is offline
    Super Member
    Nov 2008

    Beginner at definite integrals

    Could someone show me how to evaluate this:^2+%2B+2x+-+5)+dx%2C+x+%3D+1..4 please? (using right endpoints which I guess is the "default?")

    Wolfram gives me the final answer but it does not show me the precise steps. I did the work myself but I am not at home so I can't scan my work and plus my internet doesn't work at home so I can't send it either way but I got 7 as the final answer but it's supposed to be 21 according to my teacher and wolfram. Please help!

    Any help would be greatly appreciated!
    Thanks in advance!
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  2. #2
    MHF Contributor ebaines's Avatar
    Jun 2008
    First, determine what the integral is for $\displaystyle f(x) = x^2 +2x -5$:

    F(x) = \int f(x) dx = \int (x^2 +2x -5) dx = \frac {x^3} 3 + x^2 -5x + C

    Did you get that? Note that I used a lower-case f to denote the original function, and capital F to denote the integral. Next, evaluate F(x) at the point x=4, and subtract the value at x = 1:

    F(4) - F(1) = (\frac {4^3} 3 + 4^2 - 5 \cdot 4 + C) - (\frac {1^3} 3 + 1^2 -5 \cdot 1 - C) = \frac {64 - 1} 3 + 16 - 20 - 1 + 5 = 21

    Note that the constant C cancels out. Since the constant of integration always cancels when evaluating definite integrals, most people simply don't bother including it in their work, but I think it's good practice to not forget about it. Hope this helps!
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  3. #3
    Junior Member
    Sep 2009
    Just to point this out.. you can break that function down between terms...

    For f(x) = x2 + 2x -5 you can do this to integrate

    $\displaystyle \int x^2 + \int 2x - \int 5 $

    to do this you have a simple formula of $\displaystyle \int x^n = x^{n+1} / n + 1 $

    in the first case you have x^2.... following the formula this results in (x^3) / (3)

    in the case of constants such as the last 5 in the expression you simply tack on a x because when you derive 5x you get 5

    This way you can break it down and say ok I have three different indiviual integrals to break down and then you can sum up the integrals afterwards.

    Then since this is a definite integral you will get an number not a function.

    when you have f(x) properly integrated then you can go ahead and take the upperbound in this case of 4 and sub it in for x

    then you take the lower bound and sub it in for x and then you simply take your upperbound result and subtract from it your lower bound result

    hope this helps
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