How do you integrate #f(t) = 1.4e^(0.07t)#?

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Answer 1 · 2017-10-23T15:35:06

# int 1.4e^(0.07t) do = 20 e^(0.07t)#

#f(t) = int 1.4e^(0.07t)dt#

#= (7/5) inte^((7t)/100) dt#

#u = (7t)/100, du = ((7t)/100)dx, dx = (100/7)du#

Apply constant multiple rule,

#f(t) = (cancel(7/5)cancel(100/7)) 20 int e^u du#

#f(t) = 20 int e^u du = 20e^u#

But #e^u = e^(0.07)#

#:. int 1.4 e^(0.07t) dt = 20 e^(0.07t)#