With respect to the numerical evaluation of the definite integral, $K = \int_a^b x^2 \, dx$, where $a$ and $b$ are given, which of the following statements is/are TRUE? I) The value of $K$ obtained using the trapezoidal rule is always greater than or equal to the exact value of the definite integral. II) The value of $K$ obtained using the Simpson's rule is always equal to the exact value of the definite integral.

Let $f(x) = x^2$.

1. Statement I (Trapezoidal Rule): The error in the trapezoidal rule for approximating $\int_a^b f(x)\,dx$ is given by:

   $$E_T = -\frac{(b-a)^3}{12n^2} f^{\prime\prime}(\xi), \quad \text{where } \xi \in [a, b]$$

   Since $f(x) = x^2$, we have $f^{\prime\prime}(x) = 2 > 0$, which means $f(x)$ is convex (concave up). Because $f^{\prime\prime}(x) > 0$, the straight-line segments connecting points on the curve lie entirely above or on the curve, so the trapezoidal approximation always overestimates (or equals if $a = b$) the exact integral. Thus, Statement I is TRUE.

2. Statement II (Simpson's Rule): The error in Simpson's $1/3$ rule is given by:

   $$E_S = -\frac{(b-a)^5}{180n^4} f^{(4)}(\xi)$$

   Simpson's rule is exact for all polynomials of degree $\le 3$. Since $f(x) = x^2$ is a polynomial of degree 2, its fourth derivative is $f^{(4)}(x) = 0$, meaning the truncation error is identically zero. Thus, the value obtained is always equal to the exact integral. Statement II is TRUE.

Therefore, Both I and II are true, corresponding to option (C).