Work done in thermodynamic process

[![Source Jee Main 2020](https://i.stack.imgur.com/1Op9a.png)](https://i.stack.imgur.com/1Op9a.png)


Here the official answer key given by Jee main is 48 Joules which makes no sense I believe that they just blindly found the area under the curve . if I am correct the answer should be 48+2\*10(the area of rectangle below the (2,2) point ) as work done area under the PV graph with volume axis
please correct me if I am wrong




It looks like they have set the trap of shifted axis, but they have fallen for it themselves.


$$W=\int\_{V\_1}^{V\_2}{p \cdot \mathrm{d}V},$$


so $$W=(\pu{8 m3}- \pu{2 m3}) \cdot \pu{8 Pa} + (\pu{12 m3} - \pu{8 m3}) \cdot \frac{\pu{8 Pa} + \pu{2 Pa}}{2} = \pu{48 J} + \pu{20 J} = \pu{68 J}$$


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