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

1 Answer
Bill K.
Feb 24, 2016

#f(x)=e^(x)-x^2/2+3/2-e^{-1}#

Explanation:

The general antiderivative is

#\int (e^{x}-x)\ dx=e^{x}-x^{2}/2+C#

Therefore, #f(x)=e^{x}-x^{2}/2+C# for some constant #C#. Since #f(-1)=1#, we get #1=e^{-1}-(-1)^{2}/2+C=e^{-1}-1/2+C#, which implies that #C=3/2-e^{-1}#.

Hence, #f(x)=e^(x)-x^2/2+3/2-e^{-1}#.

Related questions
See all questions in Evaluating the Constant of Integration
Impact of this question
1324 views around the world
You can reuse this answer
Creative Commons License