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Therefore,
\[f(x)=\int\limits_{0}^{1}kx^2dx=k \frac{ x^3 }{ 3 }_{0}^{1}=k/3=1\]
so k=3

I have a filling I did that last step wrong.

thats correct, 0.1354*3=0.4062 is the correct probability for a)

you know how to do part b), right ?

yes that makes sense as well. Thanks.

welcome ^_^