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)`

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