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

1 Answer
georgef
Jun 24, 2016

The answer consists in calculating the primitive #f# (antiderivative) of the given function, and then evaluating the constant for which the primitive #f(2) = 3#

Explanation:

The integral distributes with respect to the sum, so:
#int (e^(x+2) + x) dx = int e^(x+2) dx + int x dx#

So, solving the first one:
#int e^(x+2) dx = int e^x e^2 dx = e^2 int e^x dx= e^2 e^x + C_1# where #C_1# is an unknown constant.

Similarly, the second one is:
#int x dx = x^2/2 + C_2#

So the antiderivative of #e^(x+2) + x# is:
#f(x) = e^2 e^x + x^2/2 + C#, where C is a constant.

Now we have to solve for #f(2) = 3#:
#e^2 e^2 + 2^2/2 + C = 3#, that is:
#e^4 + 2 + C = 3#, thus:
#e^4 + C = 1#, and then
#C = 1-e^4#.

The final answer is then:
#f(x) = e^2 e^x + x^2/2 + (1-e^4)#

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