Consider a continuous random variable X with Probability Density Function $f(x)=\dfrac{c}{(x+1)^4}$ for $x\geq0$ and $0$ otherwise.

I found $c$ to be 3 after anti-deriving $f(x)$.

Now I need to find the cumulative distribution function, $F(x)$.

Is it just $$\int_0^{ \infty}\dfrac{3}{(x+4)^4}dx$$


It would just be the integral: $$F(x)=\int_0^{x}\frac{3}{(t+1)^4}dt$$ This may be evaluated with the substitution $u=t+1$.

  • $\begingroup$ Why are the bounds from 0 to x? This is the part that confuses me $\endgroup$ – Mathgirl May 1 '15 at 1:39
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    $\begingroup$ What's the probability that $X$ is in the interval $(-\infty,x]$? $\endgroup$ – user223391 May 1 '15 at 1:47

The Cumulative Distribution, $F(x)$, is defined as $$F(x)=P(X\leq x)$$ and if you have pdf $f(x)$ this can be found by $$P(X\leq x)=\int_{-\infty}^xf(x)dx$$


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