We can do this by integrating the pdf over the given range and setting it equal to 1:
$$
\int_0^1 kx \, dx = 1
$$
Now, we can solve for k:
$$
k \int_0^1 x \, dx = 1 \\
k \left[\frac{1}{2}x^2\right]_0^1 = 1 \\
k \cdot \frac{1}{2} = 1 \\
k = 2
$$
So, the pdf
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