Integrate(ln(x)ln(1-x)dx,x=0 to 1)
Classical method, using integration by parts: \[\int\limits_{a}^{b}U \cdot dV = U \cdot V - \int\limits_{a}^{b} V \cdot dU\] But then you run into a problem... Do you see why? ;-)
This thing gets really messy fast. My recommendation? Use an approximation function. Simpon's rule will actually work, with as many decimal places of accuracy you need it depending on how large you set "n" to (n = # of sample rectangles)
Are you familiar with Simpson's Rule?
@equation do u know the answer? is it \[2-\frac{\pi^2}{6}\] ?
Actually @mukushla that is exactly equivalent to what Simpson's Rule gives, how did you shortcut that though? I would like an explanation as well for that shortcut :-)
i will post my way in few minutes
can you give the Simpson's rule
Simpson's Rule: Conditions: \(\large\frac{(b-a)}{n}\) = \(\triangle x\) n = an even #, ONLY! \[\large S_n = \frac{\triangle x}{3} \cdot [f(x_0)+4f(x_1)+2f(x_2)+...+2f(x_{n-2})+4f(x_{n-1})+f(x_n)]\]
Simpson's Rule works for ANY function, even ones that cannot be integrated. But there's sometimes exact methods that are faster and more accurate (give exact values with \(\pi\) or unusual square roots or such)
Ack it got cut off... \[S_n = \frac{\triangle x}{3} \cdot [f(x_0)+4f(x_1)+2f(x_2)+...+2f(x_{n-2})+4f(x_{n-1})+f(x_n)]\]
n is how many rectangles you are sampling by, as n goes to \(\infty\) you get the exact integral
In your case your a = 0, b = 1, and I just made n = 100 and ran it through the computer and in a second or less it had my answer.
But what the computer couldn't do was recognize could be written as 2 - \(\pi ^2\)/6
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