If \( f \) is continuous and \( \int_{0}^{16} f(x) d x=6 \), find \( \int_{0}^{4} x f\left(x^{2}\right) d x \). \( \square \)

To evaluate the integral \( \int_{0}^{4} x f(x^{2}) \, dx \), we can use the substitution method. Let's set \( u = x^{2} \). Then, we have: \[ du = 2x \, dx \quad \text{or} \quad dx = \frac{du}{2x} \] Since \( u = x^2 \), we can express \( x \) in terms of \( u \) as \( x = \sqrt{u} \). Then, substituting for \( x \) gives us \( dx = \frac{du}{2\sqrt{u}} \): When \( x = 0 \), \( u = 0 \), and when \( x = 4 \), \( u = 16 \). Now substituting into the integral: \[ \int_{0}^{4} x f(x^{2}) \, dx = \int_{0}^{16} \sqrt{u} f(u) \cdot \frac{du}{2\sqrt{u}} = \frac{1}{2} \int_{0}^{16} f(u) \, du \] This simplifies to: \[ = \frac{1}{2} \cdot 6 = 3 \] Thus, the final result is: \[ \int_{0}^{4} x f(x^{2}) \, dx = 3 \]