$$\text{Let X be a continuous random variable with pdf}$$
$$f(x) = \begin{cases} 0 & x\leq 0 \\ \frac{x}{100} e^{\frac{-x^2}{200}}& x > 0 \\ \end{cases} $$
$$\text{(a) Confirm that f(x) is a valid pdf. Hint: substitute u = } x^2/200$$ $$\text{(b) Calculate }P(X \leq 20), \text{ and } P(10 \leq X \leq 20)$$ $$\text{(c) Give an expression (not an integral!) for } P(X \leq x), \text{ the cdf of }X$$
A pdf must meet the following criteria:
$$\int_{-\infty}^{+\infty} f(x) dx = 1$$ $$ f(x) \geq 0 \ \forall x \in \mathbb{R}$$
It's trivial to show that $f(x) \geq 0$. So, taking $f(x)$ as the function above.
$$\int_{-\infty}^{+\infty} f(x) dx = \int_{-\infty}^{0} f(x) dx + \int_{0}^{+\infty} f(x) dx = \int_{0}^{+\infty} \frac{x}{100}e^{-x^2/200} dx$$
Substitute $u=-x^2/200$ and $du=-x/100$
$$-\int_{0}^{+\infty} e^u du = \left.-e^{-x^2/200}\right\rvert_{0}^{\infty} = 1 $$
The CDF can be calculated by
$$ F(x) = P( X \leq x) = \int_{-\infty}^{x} f(x) dx = \begin{cases} 0, \ \text{if} \ x \leq 0\\ 1 -e^{-x^2/200}, \ \text{if} \ x > 0 \end{cases} $$
Exercise $(b)$ is easy to solve using $F(x)$. Notice $P(10 \leq X \leq 20) = F(20) - F(10)$