If \(y = x(x^4 + x + 1)\), then we can expand the expression to get \(y = x^5 + x^2 + x\). To evaluate the definite integral \(\int \limits_{0} ^{1} y \mathrm d x\), we can find the antiderivative of \(y\) and then use the Fundamental Theorem of Calculus. The antiderivative of \(x^5 + x^2 + x\) is \(\frac{x^6}{6} + \frac{x^3}{3} + \frac{x^2}{2} + C\), where C is the constant of integration. Using the Fundamental Theorem of Calculus, we have:

\(\int \limits_{0} ^{1} y \mathrm d x = \left[\frac{x^6}{6} + \frac{x^3}{3} + \frac{x^2}{2}\right]_0^1 = \left(\frac{1^6}{6} + \frac{1^3}{3} + \frac{1^2}{2}\right) - \left(\frac{0^6}{6} + \frac{0^3}{3} + \frac{0^2}{2}\right) = \frac{1}{6} + \frac{1}{3} + \frac{1}{2} = 1\)