How do you evaluate the definite integral #int e^x# from #[0,ln2]#? Calculus Introduction to Integration Definite and indefinite integrals 1 Answer Steve M Nov 2, 2016 # int_0^ln2e^xdx = 1# Explanation: # e^x # is the only function which remains unchanged when differentiated, and consequently # inte^xdx=e^x+C # So, # int_0^ln2e^xdx = [e^x]_0^ln2 # # :. int_0^ln2e^xdx = e^ln2-e^0# # :. int_0^ln2e^xdx = 2-1# # :. int_0^ln2e^xdx = 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 4294 views around the world You can reuse this answer Creative Commons License