How do you integrate #int 2^xdx# from #[-1,2]#? Calculus Introduction to Integration Integrals of Exponential Functions 1 Answer Narad T. Dec 15, 2016 The answer is #=7/(2ln2)=5.05# Explanation: Let #u=2^x# #lnu=xln2# #u=e^(xln2)# #int2^xdx=inte^(xln2)dx=e^(xln2)/ln2=2^x/ln2# #int_-1^2 2^xdx= [2^x/ln2] _-1^2# #=1/ln2(2^2-1/2)# #=1/ln2*7/2=5.05# 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 1308 views around the world You can reuse this answer Creative Commons License