Express the following limit as a definite integral on the interval $[0,10]$.
${\lim }_{n\to \infty }{\sum }_{k=1}^n{(x_k^{\ast }-3)}^2\Delta x$

• A. ${\int }_{10}^0{(x-3)}^{2}dx$
• B. ${\int }_0^{10}{(x-3)}^{2}dx$
• C. ${\int }_0^{\infty }{(x-3)}^{2}dx$
• D. ${\int }_{\infty }^0{(x-3)}^{2}dx$

Answer: (B)

Verified step by step guidance

1. Recognize that the limit expression given is a Riemann sum, which is used to approximate the area under a curve. The expression \( \lim_{n\to\infty}\sum_{k=1}^{n}\left(x_{k}^{\ast}-3\right)^2\Delta x \) represents the sum of areas of rectangles under the curve \( (x-3)^2 \).
2. Identify the interval over which the definite integral is to be calculated. The problem specifies the interval [0, 10]. This means we are looking at the function \( (x-3)^2 \) from \( x = 0 \) to \( x = 10 \).
3. Understand that \( x_k^* \) represents sample points within each subinterval \( \Delta x \) of the partition of [0, 10]. As \( n \to \infty \), these sample points become dense, and the sum approximates the integral.
4. Translate the Riemann sum into a definite integral. The expression \( \lim_{n\to\infty}\sum_{k=1}^{n}\left(x_{k}^{\ast}-3\right)^2\Delta x \) becomes \( \int_0^{10}\left(x-3\right)^2\,dx \).
5. Set up the definite integral \( \int_0^{10}\left(x-3\right)^2\,dx \) to evaluate the area under the curve \( (x-3)^2 \) from \( x = 0 \) to \( x = 10 \). This integral represents the limit of the Riemann sum as \( n \to \infty \).