What is F(x) = int xe^(x^2-2x) +x^2 dx if F(0) = 1 ?

1 Answer
Jun 18, 2017

F(x)~=1/6 (-(3 sqrt(π) "erfi"(1 - x))/e + 2 x^3 + 3 e^((x - 2) x))+6.2285

Explanation:

I wasn't able to integrate this function using standard techniques of integration. Using the computer algebra systems of both MATLAB and WolframAlpha I was able to have the computer generate an answer.

F(x)=1/6 (-(3 sqrt(π) "erfi"(1 - x))/e + 2 x^3 + 3 e^((x - 2) x))+C

Where the term "erfi"(1-x) refers to the imaginary error function defined as

"erfi"(1-x)=2/sqrt(π) int_0^(1-x) e^(t^2) dt

At F(0)=1, we have

1/6 (-(3 sqrt(π) "erfi"(1 - 0))/e + 2 (0)^3 + 3 e^((0 - 2) 0))+C=1

1/6 (-(3 sqrt(π) "erfi"(1))/e + 3 e^(0)+C)=1

1/6 (-(3 sqrt(π) "erfi"(1))/e + 3+C)=1

-(3 sqrt(π) "erfi"(1))/e+3+C=6

-(3 sqrt(π) "erfi"(1))/e+C=3

C=3+(3 sqrt(π) "erfi"(1))/e

Plugging the right hand side into MATLAB, gives

C~~6.2285

So the anti-derivative is

F(x)~=1/6 (-(3 sqrt(π) "erfi"(1 - x))/e + 2 x^3 + 3 e^((x - 2) x))+6.2285