What is #f(x) = int e^(2x)-e^x+x dx# if #f(4 ) = 2 #?

1 Answer
Ratnaker Mehta
Jul 4, 2016

#f(x)=e^(2x)/2-e^x+x^2/2+e^4-e^8/2-6#

Explanation:

Let #I=int(e^(2x)-e^x+x)dx=e^(2x)/2-e^x+x^2/2+C#, #C# is a constant of integration.

To determine #C#, we are given the cond. that #f(4)=2#.

Now, #f(x)=I=e^(2x)/2-e^x+x^2/2+C.# Hence,
#f(4)=2 rArr e^8/2-e^4+8+C=2 rArr C=e^4-e^8/2-6. #

Sub.ing this value of #C# in #I#, we get,

#f(x)=e^(2x)/2-e^x+x^2/2+e^4-e^8/2-6#

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