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Thread: integration

  1. #1
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    Question integration

    2 times the integral (from -3 to 3) of (4-x) (sqrt(9-x^2) ) dx.

    Thank you very much.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by kittycat View Post
    2 times the integral (from -3 to 3) of (4-x) (sqrt(9-x^2) ) dx.

    Thank you very much.
    You question is:

    Evaluate:

    $\displaystyle
    \int_{-3}^3 (4-x) \sqrt{9-x^2}~dx
    $

    RonL
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by CaptainBlack View Post
    You question is:

    Evaluate:

    $\displaystyle
    \int_{-3}^3 (4-x) \sqrt{9-x^2}~dx
    $

    RonL
    $\displaystyle
    \int_{-3}^3 (4-x) \sqrt{9-x^2}~dx = \int_{-3}^3 4\sqrt{9-x^2}~dx - \int_{-3}^3 x \sqrt{9-x^2}~dx
    $

    and these two integrals can be done by the usual book methods, the first by a trig substitution and the second
    by looking at the derivative of $\displaystyle (9-x^2)^{3/2}$.

    RonL
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  4. #4
    Forum Admin topsquark's Avatar
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    $\displaystyle
    \int_{-3}^3 (4-x) \sqrt{9-x^2}~dx
    $

    Hint:
    $\displaystyle = 4 \int_{-3}^3 \sqrt{9-x^2}~dx - \int_{-3}^3 x\sqrt{9-x^2}~dx$

    You can do the second integral with a simple substitution. So let's look at the first term:
    $\displaystyle \int_{-3}^3 \sqrt{9-x^2}~dx$

    Let $\displaystyle x = 3y \implies dx = 3 dy$

    Thus
    $\displaystyle \int_{-3}^3 \sqrt{9-x^2}~dx = \int_{-1}^1 \sqrt{9-(3y)^2}~3dy$

    $\displaystyle = 3\int_{-1}^1 \sqrt{9-9y^2}~dy$

    $\displaystyle = 3\int_{-1}^1 3\sqrt{1-y^2}~dy$

    $\displaystyle = 9\int_{-1}^1 \sqrt{1-y^2}~dy$

    Does this integral look familiar?

    -Dan
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  5. #5
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    Or from the beggining you can use a simple trig. sub. defined by $\displaystyle x=3\sin u$

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  6. #6
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    In general we have that

    $\displaystyle \int\sqrt{a-x^2}\,dx=\frac12\left(x\sqrt{a-x^2}+a\arctan\frac x{\sqrt{a-x^2}}\right)+k$
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  7. #7
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    Quote Originally Posted by Krizalid View Post
    In general we have that

    $\displaystyle \int\sqrt{a-x^2}\,dx=\frac12\left(x\sqrt{a-x^2}+a\arctan\frac x{\sqrt{a-x^2}}\right)+k$
    I think you want to write $\displaystyle a^2$ (also mention that $\displaystyle a\not = 0$ , just to be safe, write $\displaystyle a>0$).
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  8. #8
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    I think you want to write $\displaystyle a^2$ (also mention that $\displaystyle a\not = 0$ , just to be safe, write $\displaystyle a>0$).
    And mention that a is a real number, and...

    -Dan
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  9. #9
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    Hi Dan,

    how do I end this one?

    9 times the integral from -1 to 1 of sqrt(1-y^2) dy - your last step! Really don't know how


    Also,
    Can anyone remind me how to do the trig substitution? I forgot it ! Please show me the steps . Thanks.
    Last edited by kittycat; Aug 17th 2007 at 01:23 PM.
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  10. #10
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    Set $\displaystyle y=\sin u\implies dy=\cos u\,du$

    Now when you have all integrated you use the first substitution to turn the trig. expressions into $\displaystyle y$ expressions.
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  11. #11
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    hi Krizalid,

    Please show me more ...! really forgot ! If possible , please show me the steps how to show the first integral by trig substitution. thank you very much.
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  12. #12
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    Change of variables according to $\displaystyle y=\sin u$ implies

    $\displaystyle \begin{aligned}
    \int {\sqrt {1 - y^2 } \,dy} &= \int {\sqrt {1 - \sin ^2 u} \cos u\,du}\\
    &= \int {\cos u\cos u\,du}\\
    &= \int {\cos ^2 u\,du}
    \end{aligned}$

    Which is easy to take.
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