How do you evaluate the definite integral #int 2e^x dx# from #[0,1]#? Calculus Introduction to Integration Definite and indefinite integrals 1 Answer Narad T. Jan 7, 2017 The answer is #=2(e-1)# Explanation: We use #inte^xdx=e^x+C# #e^0=1# Therefore, #int_0^1 2e^xdx=[2e^x]_0^1# #=(2e^1-2e^0)# #=2e-2# #=2(e-1)# Answer link Related questions What is the difference between definite and indefinite integrals? What is the integral of #ln(7x)#? Is f(x)=x^3 the only possible antiderivative of f(x)=3x^2? If not, why not? How do you find the integral of #x^2-6x+5# from the interval [0,3]? What is a double integral? What is an iterated integral? How do you evaluate the integral #1/(sqrt(49-x^2))# from 0 to #7sqrt(3/2)#? How do you integrate #f(x)=intsin(e^t)dt# between 4 to #x^2#? How do you determine the indefinite integrals? How do you integrate #x^2sqrt(x^(4)+5)#? See all questions in Definite and indefinite integrals Impact of this question 7175 views around the world You can reuse this answer Creative Commons License