Question

express the limit as a definite integral on the given interval.
lim _{n
ightarrow infty} sum_{i=1}^{n} \frac{x_{i}^{}}{left(x_{i}^{}
ight)^{2}+2} delta x, quad1,5
int_{1}^{5}left(\frac{x}{x^{2}+5}
ight) d x

Explanation:

Step1: Recall the definition of definite integral

The definite integral \(\int_{a}^{b}f(x)dx=\lim_{n
ightarrow\infty}\sum_{i = 1}^{n}f(x_{i}^*)\Delta x\), where \(a\) and \(b\) are the lower and upper limits of integration, \(x_{i}^*\) is a sample point in the \(i -\)th sub - interval \([x_{i-1},x_{i}]\), and \(\Delta x=\frac{b - a}{n}\).

Step2: Identify \(a\), \(b\) and \(f(x)\)

Given the interval \([1,5]\), so \(a = 1\) and \(b=5\).
We have \(\sum_{i = 1}^{n}\frac{x_{i}^}{(x_{i}^)^{2}+2}\Delta x\). Comparing with \(\sum_{i = 1}^{n}f(x_{i}^*)\Delta x\), we can see that \(f(x)=\frac{x}{x^{2}+2}\)

Answer:

\(\int_{1}^{5}\frac{x}{x^{2}+2}dx\)