Marginal density of X error in the test?

Question

So I took a test. And it was easy, but for some reason I got 0pts out of 5pts in a once single exercise.

OK, here it is: "Given the joint density function $$f(x,y)=12/7$$

when $$0\leq x \leq 1 \text{ and } 0\leq y \leq 1$$ And 0 elsewhere. Calculate it's marginal density of $X$."

The answers: (a): $$f_X(x)=(12x^2+6x)/7$$

(b): $$f_X(x)=(2x^2+3x)/7$$

(c): $$f_X(x)=(4+6x)/7$$

(d): $$\text{None of the above.}$$

My short solution sketch: Apply definition 5.4 (attached as image) and apply the bounds, hence $$f_X(x)=\int_0^1 12/7 dx=[12x/7]_0^1=12/7$$ thus, it's option D.

Definition 5.4 can be found here: http://puu.sh/BQaS3/5af75ea1be.png Am I right or? Because I did a similar exercise in the exercise classes where we have another number we should integrate (without variables) like here. I obtained a correct solution, but now I got 0 out of 5 points.

Answer 1

$f(x,y)$ doesn't integrate to $1$, it is not a valid joint density function.

Answer 2

I think the pdf is not right. I think the (valid) pdf is

$$f(x,y)=12/7\cdot (x^2+xy)$$

when $0\leq x \leq 1 \text{ and } 0\leq y \leq 1$

To calculate the marginal density of $X$ you have to integrate w.r.t $y$- $\color{blue}{\texttt{not x}}$

$$f_X(x)=\int_0^1 12/7\cdot (x^2+xy) \, dy $$

$$ \frac{12}{7} \cdot \left[ x^2y+\frac{1}{2}xy^2 \right]_0^1=\frac{12}7\cdot (x^2+\frac12 x)-0$$

$$=\frac{1}{7} \cdot (12x^2+6x)$$

when $0\leq x \leq 1 $

Therefore the answer is a)