Given that $\int_{0}^{1} x^{2} d x=\frac{1}{3}$, use this fact and the properties of integrals to evaluate $\int_{0}^{1}\left(7-3 x^{2}\right) d x$

Step 1 ：Given that \(\int_{0}^{1} x^{2} d x=\frac{1}{3}\), use this fact and the properties of integrals to evaluate \(\int_{0}^{1}\left(7-3 x^{2}\right) d x\)

Step 2 ：The integral of a function can be broken down into the sum of the integrals of its components. Therefore, we can break down the integral of the function \(7-3x^2\) into the integral of \(7\) and the integral of \(-3x^2\).

Step 3 ：The integral of a constant is simply the constant times the length of the interval, so the integral of \(7\) over the interval from \(0\) to \(1\) is \(7\).

Step 4 ：The integral of \(-3x^2\) can be found by multiplying the given integral of \(x^2\) by \(-3\), which gives \(-1\).

Step 5 ：Adding these two results together, we find that the integral of the function \(7-3x^2\) from \(0\) to \(1\) is \(6\).

Step 6 ：Final Answer: The integral of the function \(7-3x^2\) from 0 to 1 is \(\boxed{6}\).