How do you evaluate the definite integral #int 2e^x dx# from #[0,1]#?

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Answer 1 · 2017-01-07T05:51:04

The answer is #=2(e-1)#

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)#