How do you find the integral of #e^((-1/2)*x)#? Calculus Introduction to Integration Integrals of Exponential Functions 1 Answer Sasha P. Sep 26, 2015 #-2e^(-1/2x)+C# Explanation: #t=-1/2x => dt=-1/2dx => dx=-2dt# #I=int e^(-1/2x)dx = int e^t*(-2dt) = -2 inte^tdt = -2e^t + C# #I=-2e^(-1/2x)+C# 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 14183 views around the world You can reuse this answer Creative Commons License