0 votes 0 votes The value of $\int_{-\infty}^{+\infty} e^{-t} \delta (2t-2){d}t$, where $\delta (t)$ is the Dirac delta function, is $\dfrac{1}{2e} \\$ $\dfrac{2}{e} \\$ $\dfrac{1}{e^{2}} \\$ $\dfrac{1}{2e^{2}}$ Calculus gate2016-ee-1 calculus definite-integral + – makhdoom ghaya asked Jan 29, 2017 • retagged Mar 9, 2021 by Lakshman Bhaiya makhdoom ghaya 9.4k points answer comment Share See all 0 reply Please log in or register to add a comment.