# Joint PDF. Is my book wrong?

Consider that X and Y have the joint pdf $f(x,y) = (2/3)(x+1)$ for $0 < x < 1$ and $0 < y < 1$ and 0 otherwise. What is $P(X < 2Y < 3X)$? My book say the answer is 73/162

But I keep doing $\int_0^1 \int_{x/2}^{3x/2} f(x,y) \, dy \, dx$ and I am not getting 73/162

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The region you need to integrate is the blue region as shown in the figure below.

Hence, your integral should go as follows.

\begin{align} \int_0^{2/3} \int_{x/2}^{3x/2} f(x,y) \, dy \, dx + \int_{2/3}^1 \int_{x/2}^{1} f(x,y) \, dy \, dx & = \int_0^{2/3} \int_{x/2}^{3x/2} \dfrac23(x+1) \, dy \, dx\\ & + \int_{2/3}^1 \int_{x/2}^{1} \dfrac23 (x+1) \, dy \, dx\\ & = \int_0^{2/3} \dfrac23x(x+1) \, dx\\ & + \int_{2/3}^1 \dfrac23 (1-x/2)(x+1) \, dx\\ & = \dfrac23 \left(x^3/3+x^2/2 \right)_0^{2/3}\\ & + \dfrac23 \left(x^2/2+x - x^3/6 - x^2/4 \right)_{2/3}^1\\ & = \dfrac23 \left((2/3)^3/3+(2/3)^2/2 \right)\\ & + \dfrac23 \left(1/2+1 - 1/6 - 1/4 \right)\\ & - \dfrac23 \left((2/3)^2/2 + (2/3) \right)\\ & + \dfrac23 \left((2/3)^3/6 + (2/3)^2/4 \right) \\ & = \dfrac{73}{162} \end{align}

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Oh shoot, that graph is much much better than my quick and dirty plot on scrap paper. – Hawk Nov 2 '12 at 3:55
@jak The graph was made using grapher (en.wikipedia.org/wiki/Grapher) software on mac. – user17762 Nov 2 '12 at 4:03
I have Mathematica, so it okay. Thank you however – Hawk Nov 2 '12 at 4:08