Calculus of a Single Variable, Chapter 8, 8.6, Section 8.6, Problem 39

For the given problem: int_0^1 xe^(x^2) , we may first solve for its indefinite integral. Indefinite integral are written in the form of int f(x) dx = F(x) +C
where: f(x) as the integrand
F(x) as the anti-derivative function
C as the arbitrary constant known as constant of integration
We omit the arbitrary constant C when we have a boundary values: a to b. We follow formula: int_a^b f(x) dx = F(x)|_a^b .
Form the table of integrals, we follow the indefinite integral formula for exponential function as:
int xe^(-ax^2) dx = - 1/(2a)e^(-ax^2) +C
By comparison of -ax^2 with x^2 shows that we let a= -1 .
Plug-in a=-1 on -ax^2 for checking, we get: - (-1) x^2= +x^2 or x^2 .
Plug-in a=-1 on integral formula, we get:
int_0^1 xe^(x^2) =- 1/(2(-1))e^((-(-1)x^2))| _0^1
=- 1/(-2)e^((1*x^2))| _0^1
= 1/2e^(x^2)| _0^1
Applying definite integral formula: F(x)|_a^b = F(b)-= F(a) .
1/2e^(x^2)| _0^1 =1/2e^(1^2) -1/2e^(0^2)
=1/2e^(1) -1/2e^(0)
=1/2e -1/2 *1
= 1/2e -1/2 or 1/2(e-1)