Uniform Distribution : pdf & inverse cdf

Question

$X\sim U(1,3)$. Verify that X has cdf $F_X(x) = 2(x − 1)$ for $x \epsilon(1, 3)$ and thus that $F^{−1}_X (y) = 2y +1$ for $y \epsilon (0, 1)$.

My attempt for $F_X(x)=\int_{-\infty}^{\infty}\frac{1}{b-a}dx=\int_{1}^{x}\frac{1}{3-1}dx=\frac{1}{2}\int_{1}^{x}dx=\frac{1}{2}[x-1]$

But the result is $F_X(x) = 2(x − 1)$.

I couldn't solve $F^{−1}_X (y)$

Answer 1

Surely the question that says $F_{X}(x) = 2(x-1)$ is wrong, since if you substitute 3 for $x$ you are getting $4$ as the answer.

Answer 2

I believe that is a typo because your answer is indeed correct. For the second part you can set $x = (y-1)\frac{1}{2}$ and just solve for y. This will give you your inverse.