From science to arts, IDNLearn.com has the answers to all your questions. Ask any question and receive comprehensive, well-informed responses from our dedicated team of experts.
Sagot :
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]
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]
Thank you for joining our conversation. Don't hesitate to return anytime to find answers to your questions. Let's continue sharing knowledge and experiences! IDNLearn.com is dedicated to providing accurate answers. Thank you for visiting, and see you next time for more solutions.