Finding the area of a trapezoid given vertices I encountered this problem and did not know how to solve it:

I graphed it out - making a point for each vertice. I then decided there must be a 3 because when $x=0$ then $y=3$. But, according to that logic, it should be answer c - because when $y=0$ then $x=3$. But, that is not the answer.   
Where did I go wrong, and how should I go about solving this problem when I am not given an answer key? 
 A: When you look at the picture of the trapezoid in question, to find its area you are actually integrating a line segment from $(0,3)$ to $(3,6)$ in $dx$.
The slope of this is line is $(6-3)/(3-0) = 1$ and plugging in the first point yields the equation $y=x+3$. Thus we will integrate $(x+3)$.
As for the limits, you need to go $(0,3) \to (3,6)$ in $x$, so $x=0$ to $x=3$...
A: Well, the correct answer is c. The slope of the line that represents the upper part of the trapezoid should be $m=\frac{3}{3}=1$ (the run is $3$ and the rise is also $3$ because $6-3=3$) and the $y$-intercept of that line, $b$, is $3$. Plugging all that information into the slop-intercept form of a line $y=mx+b$ gives us:
$$
y=x+3.
$$
And to get the area under that curve, you would integrate it from $0$ to $3$:
$$
\int_{0}^{3}(x+3)\,dx.
$$
A: The equation of the segment joining $ (0,3)$ to $(3,6)$ is $$y=3+x$$
You are finding the area under   $$y=3+x$$ and over the $x$-axis from $x=0$ to $x=3$
The correct choice is $(c)$
A: The area is $\int_0^3 (x+3)dx = \frac{27}{2} = 13.5$
This is the area between the curve $y = x+3$ and the $x$-axis, between $x = 0$ and $x = 3$. You are trying to calculate this area, right?
