However, area calculated is correct but the graph is not drawn correctly.

One has to clearly specify the value of y for different values of x i.e. different interval break points such as 1/3 or 3/4 .

I agree. These are some technicalities which I assume the reader can understand from the question.

Another technicality is that the (correct)answer will be :

The Area of the function $y(x)$ is composed of an area of three rectangles, as shown in the above picture.

$\text{Total area} = \left[2\ast\left(\frac{1}{3} –h- 0 \right)\right] + \left[3\ast \left(\frac{3}{4}-h- \frac{1}{3} \right)\right] + \left[1\ast \left(1- \frac{3}{4} \right)\right]$ where $h \rightarrow 0$

But for this particular equation, the limit doesn’t affect the calculation.

$\qquad \qquad \quad = \frac{2}{3} + \frac{15}{12} + \frac{1}{4} = \frac{26}{12} = \frac{13}{6}\;\text{unit}^{2}.$