Probability density of a particle in a 2D box with given coordinates


> 
> Wavefunction of particle in a $2$D box with sides $L$ in the $x$ and $y$-axis:
>  $$φ = \sqrt{\frac {4}{L^2}} \sin\frac{n\_1πx}{L}\sin\frac{n\_2πy}{L}$$
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> For the different states ($n\_1=1$, $n\_2=2$), ($n\_1=2$, $n\_2=2$) and ($n\_1=1$, $n\_2=3$), which has the highest probability density at the coordinate ($0.25L$, $0.25L$)
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I know that the probability density is the integral of $φ^2$ but I don't know the interval for the integral. I would assume that it is $0<x<L$ but when I do this I get something completely different from the given answers. The given answers are:



> 
> For ($n\_1=2$, $n\_2=2$) $=>$ the value of the wavefunction is $\frac{4}{L^2}$
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> For ($n\_1=1$, $n\_2=2$) $=>$ the value of the wavefunction is $\frac{2}{L^2}$
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> For ($n\_1=1$, $n\_2=3$) $=>$ the value of the wavefunction is $\frac{1}{L^2}$
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> The dimension of the probability density must be $\frac{1}{length^2}$, so ($n\_1=2$, $n\_2=2$) is the correct answer.
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I don't understand how they have gotten this and why ($n\_1=2$, $n\_2=2$) is the correct answer. Could someone please explain this to me. Thank you!


