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Hard Improper integral question
- Thread starter grapz
- Start date
Hello, grapz!
I have an approach . . . does anyone want to try it?
\(\displaystyle \text{Under the radical we have: }\;-x^2+ax + bx -ab \;=\;-[x^2 - (a+b)x + ab]\)
\(\displaystyle \text{Complete the square: }\;-\left[x^2 - (a+b)x + \frac{(a+b)^2}{4} \:+\:ab - \frac{(a+b)^2}{4}\right]\)
. . \(\displaystyle = \;-\left[x^2-(a+b)x + \frac{(a+b)^2}{4}\right] + \left[\frac{(a+b)^2}{4} - ab\right] \;= \;-\left[x - \frac{a+b}{2}\right]^2 + \left[\frac{a^2+2ab + b^2}{4} - ab\right]\)
. . \(\displaystyle = \;-\left[x - \frac{a+b}{2}\right]^2 + \left[\frac{a^2-2ab + b^2}{4}\right] \;=\;-\left(x - \frac{a+b}{2}\right)^2 + \left(\frac{a-b}{2}\right)^2\)
\(\displaystyle \text{The integral becomes: }\;\int^b_a\frac{dx}{\sqrt{ (\frac{a-b}{2} )^2 - (x - \frac{a+b}{2} )^2}}\)
\(\displaystyle \text{Now let: }\:x-\frac{a+b}{2} \;=\;\frac{a-b}{2}\sin\theta\quad\hdots\)
Try it . . . you're in for a pleasant surprise . . .
\(\displaystyle a\text{ and }b\text{ are constants such that: }a < b.\)
\(\displaystyle \text{Find the value of: }\;\int^b_a \frac{dx}{\sqrt{(x - a)(b - x)}}\)
The answer is \(\displaystyle \pi\)
I have an approach . . . does anyone want to try it?
\(\displaystyle \text{Under the radical we have: }\;-x^2+ax + bx -ab \;=\;-[x^2 - (a+b)x + ab]\)
\(\displaystyle \text{Complete the square: }\;-\left[x^2 - (a+b)x + \frac{(a+b)^2}{4} \:+\:ab - \frac{(a+b)^2}{4}\right]\)
. . \(\displaystyle = \;-\left[x^2-(a+b)x + \frac{(a+b)^2}{4}\right] + \left[\frac{(a+b)^2}{4} - ab\right] \;= \;-\left[x - \frac{a+b}{2}\right]^2 + \left[\frac{a^2+2ab + b^2}{4} - ab\right]\)
. . \(\displaystyle = \;-\left[x - \frac{a+b}{2}\right]^2 + \left[\frac{a^2-2ab + b^2}{4}\right] \;=\;-\left(x - \frac{a+b}{2}\right)^2 + \left(\frac{a-b}{2}\right)^2\)
\(\displaystyle \text{The integral becomes: }\;\int^b_a\frac{dx}{\sqrt{ (\frac{a-b}{2} )^2 - (x - \frac{a+b}{2} )^2}}\)
\(\displaystyle \text{Now let: }\:x-\frac{a+b}{2} \;=\;\frac{a-b}{2}\sin\theta\quad\hdots\)
Try it . . . you're in for a pleasant surprise . . .
soroban said:
\(\displaystyle \text{The integral becomes: }\;\int^b_a\frac{dx}{\sqrt{ (\frac{a-b}{2} )^2 - (x - \frac{a+b}{2} )^2}}\)
Wow. That works out really nicely. That integral just becomes arcsin [( 2x - a - b) / ( b - a)] + c. Then solving for the improper integral is much easier. Thanks
Dr. Flim-Flam
Junior Member
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- Oct 10, 2007
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Soroban, very good. However when I plugged in the original definite integral on my TI-89, I got undefined and
when I plugged it into Maple 8, I got a bunch of goobly-gook with imaginary numbers. Does this mean some integrals (gasp)
have to be done the old fashion way?
when I plugged it into Maple 8, I got a bunch of goobly-gook with imaginary numbers. Does this mean some integrals (gasp)
have to be done the old fashion way?
Dr. Flim-Flam
Junior Member
- Joined
- Oct 10, 2007
- Messages
- 108
I get -Pi, not Pi.
Dr. Flim-Flam
Junior Member
- Joined
- Oct 10, 2007
- Messages
- 108
Soroban, does arcsin(-1) = -Pi/2 or 3Pi/2? If the first then the answer is -Pi,
if the second then the answer is Pi.
if the second then the answer is Pi.
The answer is indeed Pi. Yes, you get undefined running it through the calculator. This is an e;lliptic integral(or very close to it).
It appears to be similar to Gauss's famous \(\displaystyle \int_{-\infty}^{\infty}\frac{1}{\sqrt{(x^{2}+a^{2})+(x^{2}+b^{2})}}dx=\frac{\pi}{agm(a,b)}\). Where agm(a,b) is the arithmetic-geometric mean.
From which one gets \(\displaystyle \int_{-\infty}^{\infty}\frac{1}{x^{2}+(agm(a,b))^{2}}dx=\frac{\pi}{agm(a,b)}\)
It appears to be similar to Gauss's famous \(\displaystyle \int_{-\infty}^{\infty}\frac{1}{\sqrt{(x^{2}+a^{2})+(x^{2}+b^{2})}}dx=\frac{\pi}{agm(a,b)}\). Where agm(a,b) is the arithmetic-geometric mean.
From which one gets \(\displaystyle \int_{-\infty}^{\infty}\frac{1}{x^{2}+(agm(a,b))^{2}}dx=\frac{\pi}{agm(a,b)}\)
Dr. Flim-Flam
Junior Member
- Joined
- Oct 10, 2007
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- 108
galactus, I think I found a contradiction on my TI-89. The answer comes out to be arcsin(-1)-arcsin(1) = -Pi
However if I plug constants into the original integral, a<b on my TI-89, I will get Pi. However when I punch in arcsin(-1) into my
trusty TI-89, I get -PI/2, not 3Pi/2. Hence the TI-89 uses 3PI/2 to give Pi to the definite integral but gives -pi/2 for
arcsin(-1).
However if I plug constants into the original integral, a<b on my TI-89, I will get Pi. However when I punch in arcsin(-1) into my
trusty TI-89, I get -PI/2, not 3Pi/2. Hence the TI-89 uses 3PI/2 to give Pi to the definite integral but gives -pi/2 for
arcsin(-1).