GATE 2015 EE – SET 2 – Question 26

Solution :

V = $$\int _{y=0}^{1}$$ $$e^{x}$$ $$\left | _{0}^{y}$$dy

$$\int _{y=0}^{1} (e^{y}-1)$$  dy =$$ (e^{y}-y)_{0}^{1} $$= $$e^{1}-1-e^{0}$$ = e -2