How do you integrate #int 2^xdx# from #[-1,2]#?

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Answer 1 · 2016-12-15T14:11:55

The answer is #=7/(2ln2)=5.05#

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#