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

Question

7281 views around the world

You can reuse this answer
Creative Commons License

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)#