Find $a$ so that $a(e^{-2x}-e^{-3x})$ is a probability density function.

Question

Let $f(x) = a(e^{-2x}-e^{-3x}),$ for $x\geq 0$, and $f(x) = 0$ elsewhere.
(a) Find $a$ so that $f(x)$ is a probability density function.
(b) What is $P(X\leq 1)$?

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If it is possible, please show me the detail steps how to solve the first question.

Thank you very very much!

Answer 1

Hint: $a)$ to be a density function, it must be : $\displaystyle \int_{-\infty}^{\infty} f(x)dx = 1$. Split the integral.

$b)$ $P(X\leq 1) = \displaystyle \int_{-\infty}^1 f(x)dx= \displaystyle \int_{0}^1 f(x)dx$

Answer 2

$a$ is determined by the equation: $$\int_{-\infty}^\infty f(x) \ dx = 1$$So for this problem:$$a\int_{0}^\infty e^{-2x}-e^{-3x}\ dx = 1$$Now you can solve for $a$.
For the second question, $$P(X\le 1)=\int_{0}^1 f(x) \ dx$$ Plug the values in and you can solve for the probability.