We are asked to evaluate the definite integral $I = \int_{0}^{4} xe^{x^2} dx$.

1. Problem Description

We are asked to evaluate the definite integral $I = \int_{0}^{4} xe^{x^2} dx$.

2. Solution Steps

To solve this integral, we can use u-substitution. Let $u = x^2$. Then, $du = 2x \, dx$. This means $x \, dx = \frac{1}{2} du$.

When $x = 0$, we have $u = 0^2 = 0$.

When $x = 4$, we have $u = 4^2 = 16$.

Now we can rewrite the integral in terms of $u$:

$I = \int_{0}^{16} e^u \frac{1}{2} du = \frac{1}{2} \int_{0}^{16} e^u du$.

The integral of $e^u$ with respect to $u$ is $e^u$. So,

$I = \frac{1}{2} [e^u]_{0}^{16} = \frac{1}{2} (e^{16} - e^0)$.

Since $e^0 = 1$, we have

$I = \frac{1}{2} (e^{16} - 1)$.

3. Final Answer

$\frac{1}{2}(e^{16} - 1)$