0 votes 0 votes The value of the integral $\int_{0}^{2}\int_{0}^{x}e^{x+y}dydx$ is $\dfrac{1}{2}(e-1) \\$ $\dfrac{1}{2}(e^2-1)^2 \\$ $\dfrac{1}{2}(e^2-e) \\$ $\dfrac{1}{2}(e-\frac{1}{e})^2$ Calculus gateme-2014-set4 calculus definite-integrals + – Arjun asked Feb 19, 2017 • recategorized Mar 4, 2021 by Lakshman Bhaiya ♦Arjun 28.5k points answer See all 0 reply