How do you integrate #int e^(-2x)dx# from #[0,1]#? Calculus Introduction to Integration Integrals of Exponential Functions 1 Answer Andrea S. Nov 28, 2016 In general #int(e^(alphax))dx = 1/alphae^(alphax)# Explanation: #int_0^1e^(-2x)dx = -1/2e^(-2x)|_0^1=-1/2 (e^-2-1)# Answer link Related questions How do you evaluate the integral #inte^(4x) dx#? How do you evaluate the integral #inte^(-x) dx#? How do you evaluate the integral #int3^(x) dx#? How do you evaluate the integral #int3e^(x)-5e^(2x) dx#? How do you evaluate the integral #int10^(-x) dx#? What is the integral of #e^(x^3)#? What is the integral of #e^(0.5x)#? What is the integral of #e^(2x)#? What is the integral of #e^(7x)#? What is the integral of #2e^(2x)#? See all questions in Integrals of Exponential Functions Impact of this question 27726 views around the world You can reuse this answer Creative Commons License