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Thread: Integrals

  1. #1
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    Integrals

    Could somebody please explain how i can integrate these, i have tried but i just get lost.

    1)$\displaystyle \int(2+0.5x-6x^2)dx$

    2) $\displaystyle \int(x-x^{-2}-x^{0.5})dx$
    Last edited by ForgottenMemorie; Mar 27th 2008 at 03:45 AM. Reason: Formatting
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  2. #2
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    Quote Originally Posted by ForgottenMemorie View Post
    $\displaystyle \int(2+0.5x-6x^2)dx$

    $\displaystyle \int (x - x^{-2} - x^{0.5}) dx$ Mr F asks: Is this what you meant?
    For both integrals, break them up into seperate integrals and use the rule $\displaystyle \int x^n \, dx = \frac{1}{n+1} x^{n+1}, \, n \neq -1$.
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    Quote Originally Posted by mr fantastic View Post
    For both integrals, break them up into seperate integrals and use the rule $\displaystyle \int x^n \, dx = \frac{1}{n+1} x^{n+1}, \, n \neq -1$.
    For example, for question 1: $\displaystyle \int x^2 \, dx = \frac{1}{2+1} x^{2+1},$
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    Quote Originally Posted by ForgottenMemorie View Post
    For example, for question 1: $\displaystyle \int x^2 \, dx = \frac{1}{2+1} x^{2+1},$ Mr F adds: $\displaystyle {\color{red} = \frac{1}{3} x^3}$.
    Hmmm well yes, I guess ....... in the sense that

    $\displaystyle \int 2 + 0.5x - 6x^2 \, dx = \int 2 \, dx + 0.5 \int x \, dx - 6 \int x^2 \, dx$

    and so you've done the last integral bit .....
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  5. #5
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    I dont know how to do it, need help please.
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  6. #6
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    Quote Originally Posted by ForgottenMemorie View Post
    Could somebody please explain how i can integrate these, i have tried but i just get lost.

    1)$\displaystyle \int(2+0.5x-6x^2)dx$

    2) $\displaystyle \int(x-x^{-2}-x^{0.5})dx$
    $\displaystyle \int{2+\frac{x}{2}-6x^2}dx$...so using the general rule you get $\displaystyle 2x+frac{x^2}{4}-2x^3+C$

    and using the same method Mr. fantastic showed you you get the antiderivative of the other to be $\displaystyle \frac{x^2}{2}+\frac{1}{x}-\frac{2x^{\frac{3}{2}}}{3}+C$
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  7. #7
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    Hey,

    I'm in Calc I right now, And I learned how to do this, so:

    There's two rules to keep in mind when you work these:

    Constant rule: You can kick a constant out of the integral to make it easier:

    $\displaystyle \int af(x) \, dx = a\int f(x) \, dx $

    And, you can break it up and integrate the smaller parts, for example:

    $\displaystyle \int f(x) + g(x) + h(x) \, dx = \int f(x) \, dx + \int g(x) \, dx + \int h(x) \, dx $

    And remember the antiderivative power rule that Mr. F presented:

    $\displaystyle
    \int x^n \, dx = \frac{1}{n+1} x^{n+1}, \, n \neq -1
    $

    for example:

    Integrate:
    $\displaystyle \int 2x^2 + 4x + 3 \, dx$

    $\displaystyle \int 2x^2 \, dx + \int 4x \, dx + \int 3 \, dx$
    I broke it up into the integration of separate terms.

    $\displaystyle 2\int x^2 \, dx + 4\int x \, dx + \int 3 \, dx$
    Then I isolated the constant in each term.

    Now we begin integration
    $\displaystyle 2 (\frac {1}{2+1} x^{2+1}) + 4 (\frac {1}{1+1} x^{1+1}) + 3x + C$
    Here, we used the antiderivative power rule, as described by Mr. F and this simplifies to:

    $\displaystyle \frac {2}{3}x^3 + 2x^2 + 3x + C$

    And that's the integral.

    Also, think in terms of fractions $\displaystyle 0.5 = \frac{1}{2}$ for the second problem, it will help enormously. And don't forget the plus C! very important.

    Hope this helps!
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