1. ## 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?

2. ## Re: Integral

Originally Posted by Stuck Man
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.

3. ## Re: Integral

Originally Posted by Stuck Man
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*}

4. ## 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.

5. ## 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.

6. ## Re: Integral

Originally Posted by Stuck Man
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.

7. ## 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.

8. ## Re: Integral

Originally Posted by Stuck Man
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

9. ## 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

10. ## Re: Integral

Originally Posted by Mattj
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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# integral of root(x-a)(b-x)

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