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Damn, this integration looks fishy! Integrate with respect to y? WTF?



Anyone have any idea what this is? Never seen it before and the evening before the exam practicing another past paper I see this.

I only know how to integrate x, but not f(x)

What is that and how do I solve it?

Thanks.
(edited 12 years ago)
P(AB)=P(A)P(B)\mathrm{P}(A \cap B) = \mathrm{P}(A)\mathrm{P}(B) if and only if A and B are independent. This is why you get the result that you do, since P(A|B) is just P(A) since A won't depend on B.

Is it specified that they're independent?
Original post by in_your_face
when we have conditional probability, P(A|B) = P(AnB)/P(B), right?

but if AnB= A times B then P(A|B) = P(A)xP(B)/P(B) and just equal to P(A), which is not right, since I don't get the right answer.

Where am I wrong?

Thanks.


What Wanderlust said. P(AnB) is not necessarily P(A)xP(B).
Reply 3
thanks guys.

integration suggestions anyone?
(edited 12 years ago)
Reply 4
Original post by in_your_face
thanks guys.

integration suggestions anyone?


For the first part, you can take constants outside the integral, so you can divide both sides by 3. Swapping the limits results in multiplying the integral by -1, giving the required result. For the second part, you can split the integrals up into an integral you know how to integrate, and the integral given in the question.
Original post by in_your_face
thanks guys.

integration suggestions anyone?

For the second one, note that abydx+bcydx=acydx\displaystyle\int ^b_a y dx + \displaystyle\int ^c_b y dx = \displaystyle\int ^c_a y dx (Where y is a continuous function of x and is integrable over [a,c][a,c]).
(edited 12 years ago)
Reply 6
Original post by Phil_Waite
For the first part, you can take constants outside the integral, so you can divide both sides by 3. Swapping the limits results in multiplying the integral by -1, giving the required result. For the second part, you can split the integrals up into an integral you know how to integrate, and the integral given in the question.



Original post by Farhan.Hanif93
For the second one, note that abydx+bcydx=acydx\displaystyle\int ^b_a y dx + \displaystyle\int ^c_b y dx = \displaystyle\int ^c_a y dx (Where y is a continuous function of x and is integrable over [a,c][a,c]).


Many thanks!
Can't believe I wasn't taught that. >:frown:

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