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Math Help - Integral

  1. #1
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    Integral

    With this question do I simply replace x in the integral, simplify and integrate?

    "Using the substitution x=(a+b)/2 +((b-a)/2)cos t, evaluate Integral from a to b of square root of((x-a)(b-x))dx".

    Do I need new limits?
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  2. #2
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    Re: Integral

    Quote Originally Posted by Stuck Man View Post
    With this question do I simply replace x in the integral, simplify and integrate?

    "Using the substitution x=(a+b)/2 +((b-a)/2)cos t, evaluate Integral from a to b of square root of((x-a)(b-x))dx".

    Do I need new limits?
    Of course! You substitute for x, dx and the integral terminals.
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  3. #3
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    Re: Integral

    Quote Originally Posted by Stuck Man View Post
    With this question do I simply replace x in the integral, simplify and integrate?

    "Using the substitution x=(a+b)/2 +((b-a)/2)cos t, evaluate Integral from a to b of square root of((x-a)(b-x))dx".

    Do I need new limits?
    You have \displaystyle \begin{align*}\int_a^b{\sqrt{(x-a)(b-x)}\,dx} &= \int_a^b{\sqrt{-x^2 + (a+b)x - ab}\,dx} \\ &= \int_a^b{\sqrt{-\left[x^2 - (a+b)x + ab\right]}\,dx} \\ &= \int_a^b{\sqrt{-\left\{x^2 - (a + b)x + \left[-\left(\frac{a+b}{2}\right)\right]^2 - \left[-\left(\frac{a+b}{2}\right)\right]^2 + ab\right\}}\,dx} \\ &= \int_a^b{\sqrt{-\left[\left(x - \frac{a+b}{2}\right)^2 - \frac{a^2 - 2ab + b^2}{4}\right]}\,dx} \\ &= \int_a^b{\sqrt{\left(\frac{a - b}{2}\right)^2 - \left(x - \frac{a+b}{2}\right)^2}\,dx}\end{align*}

    So now make the substitution \displaystyle x - \frac{a+b}{2} = \left(\frac{a-b}{2}\right)\cos{\theta} \implies dx = -\left(\frac{a-b}{2}\right)\sin{\theta}\,d\theta ,

    also noting that when \displaystyle x = a

    \begin{align*}a - \frac{a+b}{2} &= \left(\frac{a-b}{2}\right)\cos{\theta} \\ \frac{a-b}{2} &= \left(\frac{a-b}{2}\right)\cos{\theta} \\ 1 &= \cos{\theta} \\ \theta &= 0\end{align*}

    and when \displaystyle x = b

    \displaystyle \begin{align*} b - \frac{a+b}{2} &= \left(\frac{a-b}{2}\right)\cos{\theta} \\ -\left(\frac{a-b}{2}\right) &= \left(\frac{a-b}{2}\right)\cos{\theta} \\ -1 &= \cos{\theta} \\ \theta &= \pi \end{align*}

    So the integral becomes

    \displaystyle \begin{align*} \int_a^b{\sqrt{\left(\frac{a - b}{2}\right)^2 - \left(x - \frac{a+b}{2}\right)^2}\,dx} &= \int_0^{\pi}{\sqrt{\left(\frac{a - b}{2}\right)^2 - \left[\left(\frac{a-b}{2}\right)\cos{\theta}\right]^2}\left[-\left(\frac{a-b}{2}\right)\sin{\theta}\right]\,d\theta} \\ &= -\left(\frac{a-b}{2}\right)\int_0^{\pi}{\sin{\theta}\, \sqrt{ \left( \frac{a-b}{2} \right)^2 - \left( \frac{a-b}{2} \right)^2 \cos^2{\theta} } \, d\theta} \\ &= -\left(\frac{a-b}{2}\right) \int_0^{\pi}{\sin{\theta}\, \sqrt{ \left( \frac{a-b}{2} \right)^2\left( 1 - \cos^2{\theta} \right)  } \,d\theta } \\ &= -\left( \frac{a-b}{2} \right) \int_0^{\pi}{ \sin{\theta} \, \sqrt{ \left( \frac{a-b}{2} \right)^2 \sin^2{\theta} } \,d\theta} \\ &= -\left(\frac{a-b}{2}\right) \int_0^{\pi}{\sin{\theta} \left( \frac{a-b}{2} \right) \sin{\theta}\,d\theta} \\ &= -\left(\frac{a-b}{2}\right)^2 \int_0^{\pi}{ \sin^2{\theta} \,d\theta }\end{align*}

    \begin{align*} &= -\left(\frac{a-b}{2}\right)^2 \int_0^{\pi}{ \frac{1}{2} - \frac{1}{2}\cos{2\theta} \,d\theta} \\ &= -\left(\frac{a-b}{2}\right)^2 \left[\frac{1}{2}\theta + \frac{1}{4}\sin{2\theta} \right]_0^{\pi} \\ &= -\left(\frac{a-b}{2}\right)^2 \left[\left(\frac{\pi}{2} + \frac{1}{4}\sin{2\pi}\right) - \left(\frac{0}{2} + \frac{1}{4}\sin{0}\right)\right] \\ &= -\frac{\pi}{2}\left(\frac{a-b}{2}\right)^2 \\ &= -\frac{\pi (a-b)^2}{8}\end{align*}
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    Re: Integral

    Geometric interpretation: the integrand function is y=\sqrt{(x-a)(b-x)} . Taking squares we see that y represents a circle centered ((a+b)/2,0) and radius r=\sqrt{(a-b)^2/4} . So the integral I is half of the area of the corresponding disk, that is I=(1/2)\pi (a-b)^2/4}=\pi (a-b)^2/8 .

    P.S. Prove It: surely you have a computation mistake, the integral must be positive.
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    Re: Integral

    Prove It, you have not used the value of x given exactly. That is the first mistake. Also near the end cos has been integrated to -sin. I have tried to get the answer but I ended with the same answer. I too think it should be positive.
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    Re: Integral

    Quote Originally Posted by Stuck Man View Post
    Prove It, you have not used the value of x given exactly. That is the first mistake. Also near the end cos has been integrated to -sin. I have tried to get the answer but I ended with the same answer. I too think it should be positive.
    It shouldn't make any difference if you substitute \displaystyle \left(\frac{a-b}{2}\right)\cos{\theta} because \displaystyle \left(\frac{a-b}{2}\right)\cos{\theta} = -\left(\frac{b-a}{2}\right)\cos{\theta}, and when you make the substitution, it gets squared anyway, giving the same value.
    You are correct that \displaystyle \frac{1}{2}\cos{2\theta} should integrate to \displaystyle \frac{1}{4}\sin{2\theta}, but that sign error doesn't make any difference since it goes to 0 from substituting both terminals anyway.

    I have found what my mistake is. Since \displaystyle b \geq a, that means when I took the square root, I should have done \displaystyle \sqrt{\left(\frac{a-b}{2}\right)^2} = -\left(\frac{a-b}{2}\right), not \displaystyle \frac{a-b}{2} to get the positive value.
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    Re: Integral

    The radius of the semicircle can be simplified to r=b-a. The limits are the opposite to those Prove It found due to misreading the question. That makes the integral positive and it all looks ok to me now. Thanks everybody.
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  8. #8
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    Re: Integral

    Quote Originally Posted by Stuck Man View Post
    The radius of the semicircle can be simplified to r=b-a. The limits are the opposite to those Prove It found due to misreading the question. That makes the integral positive and it all looks ok to me now. Thanks everybody.
    I didn't misread anything, my method was perfectly valid
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    Re: Integral

    Posting another problem which is similar in nature.Appreciate your help in solving this
    How to prove that I = integral sq. root of [(x-a)(b-x)]dx = 1/8pi(b-a)^2

    a much better view of this above problem is in the attached doc file
    Attached Files Attached Files
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    Re: Integral

    Quote Originally Posted by Mattj View Post
    Posting another problem which is similar in nature.Appreciate your help in solving this How to prove that I = integral sq. root of [(x-a)(b-x)]dx = 1/8pi(b-a)^2 a much better view of this above problem is in the attached doc file
    Hint : The integral in the attached document is I_1=\int_a^b\sqrt{(x-a)(b-x)}\;dx and the OP's integral is I_2=\int_a^b\sqrt{(x-a)(b-x)}\;dx . Find a relation between I_1 and I_2 .
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