Question

1. find a definite integral that is equal to $limlimits_{n

ightarrowinfty}sumlimits_{i = 0}^{n - 1}(2+\frac{i}{n})^{2}cdot(\frac{1}{n})$.

Explanation:

Step1: Recall the definition of definite integral

The definite integral \(\int_{a}^{b}f(x)dx=\lim_{n
ightarrow\infty}\sum_{i = 0}^{n - 1}f(x_{i})\Delta x\), where \(\Delta x=\frac{b - a}{n}\) and \(x_{i}=a + i\Delta x\).

Step2: Identify \(\Delta x\) and \(x_{i}\)

Given \(\Delta x=\frac{1}{n}\), so \(b - a=1\). And \(x_{i}=2+\frac{i}{n}\), comparing with \(x_{i}=a + i\Delta x\), we can see that \(a = 2\), then \(b=a+\Delta x\times n=2 + 1=3\).

Step3: Identify \(f(x)\)

Since \(f(x_{i})=(2+\frac{i}{n})^{2}\), and \(x_{i}=2+\frac{i}{n}\), then \(f(x)=x^{2}\).

Answer:

\(\int_{2}^{3}x^{2}dx\)