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

in_your_face
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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.
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wanderlust.xx
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\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?
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ok_cub2008
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(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).
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in_your_face
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thanks guys.

integration suggestions anyone?
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Phil_Waite
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(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.
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Farhan.Hanif93
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(Original post by in_your_face)
thanks guys.

integration suggestions anyone?
For the second one, note that \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]).
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in_your_face
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(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 \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]).
Many thanks!
Can't believe I wasn't taught that. >
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