How do you evaluate the definite integral #int 2^x dx# from #[-1,1]#?

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Answer 1 · 2016-12-31T14:15:30

#3/(2ln2)#

let's start by differentiating #" "2^x#.

#y=2^x=>lny=xln2#

#1/y(dy)/(dx)=ln2=>(dy)/(dx)=yln2=2^xln2#

so remembering integration is the reverse of differentiating.

#y=2^x=>(dy)/(dx)=2^xln2#

#int2^xdx=2^x/ln2+C#

#2^x>0, AA x inRR#, there is no negative area so we can

insert the limits and evaluate immediately.

#=1/ln2[2^x]_(-1)^(1)#

#=1/ln2(2^1-2^(-1))#

#=3/(2ln2)#