Use this definition with right endpoints to find an expression for the area under the graph of [tex]\( f \)[/tex] as a limit. Do not evaluate the limit.

[tex]\[
f(x) = x \sqrt{x^3 + 9}, \quad 1 \leq x \leq 5
\][/tex]

[tex]\[
A = \lim_{n \rightarrow \infty} \sum_{i=1}^n \square
\][/tex]



Answer :

To find an expression for the area under the graph of [tex]\( f(x) = x \sqrt{x^3 + 9} \)[/tex] from [tex]\( x = 1 \)[/tex] to [tex]\( x = 5 \)[/tex] using right endpoints in a Riemann sum, follow these steps:

1. Partition the Interval: Divide the interval [tex]\([1, 5]\)[/tex] into [tex]\( n \)[/tex] subintervals of equal width. The width [tex]\(\Delta x\)[/tex] of each subinterval is given by:
[tex]\[ \Delta x = \frac{5 - 1}{n} = \frac{4}{n} \][/tex]

2. Determine Right Endpoints: The right endpoint of the [tex]\( i \)[/tex]-th subinterval is:
[tex]\[ x_i = 1 + i \Delta x \][/tex]
where [tex]\( i \)[/tex] ranges from 1 to [tex]\( n \)[/tex].

3. Evaluate the Function at Right Endpoints: Evaluate the function [tex]\( f(x) = x \sqrt{x^3 + 9} \)[/tex] at the right endpoints [tex]\( x_i \)[/tex]:
[tex]\[ f(x_i) = \left( 1 + i \frac{4}{n} \right) \sqrt{\left(1 + i \frac{4}{n} \right)^3 + 9} \][/tex]

4. Set Up the Riemann Sum: The Riemann sum using right endpoints for the area under [tex]\( f(x) \)[/tex] from [tex]\( x = 1 \)[/tex] to [tex]\( x = 5 \)[/tex] is:
[tex]\[ \sum_{i=1}^n f(x_i) \Delta x \][/tex]
Substituting our expressions for [tex]\( f(x_i) \)[/tex] and [tex]\(\Delta x\)[/tex]:
[tex]\[ \sum_{i=1}^n \left( 1 + i \frac{4}{n} \right) \sqrt{\left(1 + i \frac{4}{n} \right)^3 + 9} \cdot \frac{4}{n} \][/tex]

5. Express the Area as a Limit: Take the limit of the Riemann sum as the number of subintervals [tex]\( n \)[/tex] approaches infinity:
[tex]\[ A = \lim_{n \to \infty} \sum_{i=1}^n \left( 1 + i \frac{4}{n} \right) \sqrt{\left(1 + i \frac{4}{n} \right)^3 + 9} \cdot \frac{4}{n} \][/tex]

Thus, the expression for the area under the graph of [tex]\( f(x) \)[/tex] as a limit is:
[tex]\[ A = \lim_{n \to \infty} \sum_{i=1}^n \left( 1 + i \frac{4}{n} \right) \sqrt{\left(1 + i \frac{4}{n} \right)^3 + 9} \cdot \frac{4}{n} \][/tex]