Question

exprese como una integral definida: $limlimits_{n \to +infty} sumlimits_{i=1}^{n} \frac{8i^2}{n^3}$

Explanation:

Step1: Recall the definition of definite integral

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

Step2: Analyze the given sum

We have \( \lim_{n
ightarrow+\infty}\sum_{i = 1}^{n}\frac{8i^{2}}{n^{3}} \). Let's rewrite the general term \( \frac{8i^{2}}{n^{3}} \). We can express it as \( 8\cdot\frac{i^{2}}{n^{2}}\cdot\frac{1}{n} \). Let's assume \( \Delta x=\frac{1}{n} \) (so that \( b - a = 1 \), and if we take \( a = 0 \), then \( b=1 \)) and \( x_{i}=i\Delta x=\frac{i}{n} \). Then \( f(x_{i}) = 8x_{i}^{2} \), since \( \frac{i^{2}}{n^{2}}=x_{i}^{2} \) and \( \frac{1}{n}=\Delta x \).

Step3: Identify the function and limits

Comparing with the definition \( \lim_{n
ightarrow+\infty}\sum_{i = 1}^{n}f(x_{i})\Delta x \), we have \( f(x)=8x^{2} \), \( a = 0 \), \( b = 1 \) and \( \Delta x=\frac{1}{n} \), \( x_{i}=\frac{i}{n} \). So the limit can be written as the definite integral \( \int_{0}^{1}8x^{2}dx \).

Answer:

\(\int_{0}^{1}8x^{2}dx\)