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How do you evaluate the definite integral #int e^x# from #[0,ln2]#?

1 Answer

Nov 2, 2016

# int_0^ln2e^xdx = 1#

Explanation:

# e^x # is the only function which remains unchanged when differentiated, and consequently # inte^xdx=e^x+C #

So, # int_0^ln2e^xdx = [e^x]_0^ln2 #
# :. int_0^ln2e^xdx = e^ln2-e^0#
# :. int_0^ln2e^xdx = 2-1#
# :. int_0^ln2e^xdx = 1#

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