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