What is #f(x) = int e^(5x-1)+x dx# if #f(2) = 3 #?

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Answer 1 · 2017-07-06T07:45:18

#f(x)=1/5(e^(5x-1)+5-e^9)#

#f(x)=int(e^(5x-1)+x)dx" "f(2)=3#

now
#d/(dx)(e^(5x-1))=5e^(5x-1)#

using the chain rule. (This is left as an exercise for the reader to show)

by inspection therefore we have

#f(x)=int(e^(5x-1)+x)dx=1/5e^(5x-1)+1/2x^2+C#

but #f(2)=3#

#:.3=1/5e^((5xx2-1))+1/2xx2^2+C#

#3=1/5e^9+2+C#

#15=e^9+10+5C#

#:.C=(5-e^9)/5#

#f(x)=1/5e^(5x-1)+(5-e^9)/5#

#f(x)=1/5(e^(5x-1)+5-e^9)#