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Math Help - 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:

    <br />
\int_{-3}^3 (4-x) \sqrt{9-x^2}~dx<br />

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

    Evaluate:

    <br />
\int_{-3}^3 (4-x) \sqrt{9-x^2}~dx<br />

    RonL
    <br />
\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 <br />

    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 (9-x^2)^{3/2}.

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

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

    Let x = 3y \implies dx = 3 dy

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

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

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

    = 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 x=3\sin u

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  6. #6
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    In general we have that

    \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

    \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 a^2 (also mention that a\not = 0 , just to be safe, write 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 a^2 (also mention that a\not = 0 , just to be safe, write 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; August 17th 2007 at 02:23 PM.
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  10. #10
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    Set 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 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 y=\sin u implies

    \begin{aligned}<br />
\int {\sqrt {1 - y^2 } \,dy} &= \int {\sqrt {1 - \sin ^2 u} \cos u\,du}\\<br />
&= \int {\cos u\cos u\,du}\\<br />
&= \int {\cos ^2 u\,du}<br />
\end{aligned}

    Which is easy to take.
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