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Suppose that X is a continuous Random variable with probability density function given by

$$ f(x) = x^2 + \frac{2}{3}x + \frac{1}{3} \text{ for } 0 \leq x \leq c $$

What must be the value of c? And why?

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    $\begingroup$ The integral of the probability density is required to be $1$. So take that integral for arbitrary $c$ and set it equal to $1$. $\endgroup$ – Ian Oct 19 '16 at 14:11
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Hint: you want the probability of anything happening equal to one. So therefore $$ \int_{0}^{c} x^2 + \frac{2}{3}x + \frac{1}{3} = 1. $$

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