Derivation of the Right Riemann Sum Formula

Oct 13, 2025 by recoveryshake.com 53 views

Let $f$ be the function given by $f(x) = \frac{x^2 + 1}{\sqrt{x} + x + 5}$

In this article, we will explore the concept of right Riemann sum approximation for a given function $f(x) = \frac{x^2 + 1}{\sqrt{x} + x + 5}$ , which is known to be increasing on the interval $[1, 7]$ . The right Riemann sum is a method used to approximate the value of a definite integral. We will derive the formula for the right Riemann sum and then apply it to the given function.

The Function $f(x)$

The function $f(x)$ is given by the equation $f(x) = \frac{x^2 + 1}{\sqrt{x} + x + 5}$ . This function is defined for all $x \geq 0$ , and it is known to be increasing on the interval $[1, 7]$ .

Increasing Function

A function $f(x)$ is said to be increasing on an interval $[a, b]$ if for any two points $x_1$ and $x_2$ in the interval, $f(x_1) \leq f(x_2)$ whenever $x_1 \leq x_2$ . In other words, the function is increasing if its value increases as the input increases.

Right Riemann Sum

The right Riemann sum is a method used to approximate the value of a definite integral. It is defined as follows:

R_n = \sum_{i=1}^n f(x_i) \Delta x

where $x_i$ is the right endpoint of the $i$ th subinterval, $\Delta x$ is the width of each subinterval, and $n$ is the number of subintervals.

Derivation of the Right Riemann Sum Formula

To derive the formula for the right Riemann sum, we start by dividing the interval $[a, b]$ into $n$ equal subintervals. Let $x_i$ be the right endpoint of the $i$ th subinterval, and let $\Delta x$ be the width of each subinterval. Then, the right Riemann sum is given by:

R_n = \sum_{i=1}^n f(x_i) \Delta x

We can rewrite this formula as:

R_n = \sum_{i=1}^n f(x_i) \frac{b-a}{n}

where $b-a$ is the length of the interval $[a, b]$ .

Applying the Right Riemann Sum to the Function $f(x)$

Now, we will apply the right Riemann sum to the function $f(x) = \frac{x^2 + 1}{\sqrt{x} + x + 5}$ . We will divide the interval $[1, 7]$ into $n$ equal subintervals, and we will use the right endpoint of each subinterval as the input to the function.

Let $x_i$ be the right endpoint of the $i$ th subinterval, and let $\Delta x$ be the width of each subinterval. Then, the right Riemann sum is given by:

R_n = \sum_{i=1}^n f(x_i) \Delta x

We can rewrite this formula as:

R_n = \sum_{i=1}^n \frac{x_i^2 + 1}{\sqrt{x_i} + x_i + 5} \frac{7-1}{n}

where $7-1$ is the length of the interval $[1, 7]$ .

Computing the Right Riemann Sum

To compute the right Riemann sum, we need to evaluate the function $f(x)$ at the right endpoint of each subinterval. We will use the following values for $x_i$ :

$i$	$x_i$
1	1.5
2	2.5
3	3.5
4	4.5
5	5.5
6	6.5
7	7

We will evaluate the function $f(x)$ at each of these values and then compute the right Riemann sum.

Evaluating the Function $f(x)$

We will evaluate the function $f(x)$ at each of the values $x_i$ :

$i$	$x_i$	$f(x_i)$
1	1.5	2.25
2	2.5	3.25
3	3.5	4.25
4	4.5	5.25
5	5.5	6.25
6	6.5	7.25
7	7	8.25

Computing the Right Riemann Sum

Now, we will compute the right Riemann sum using the values of $f(x_i)$ that we obtained in the previous step:

R_n = \sum_{i=1}^n f(x_i) \Delta x

We can rewrite this formula as:

R_n = \sum_{i=1}^n \frac{x_i^2 + 1}{\sqrt{x_i} + x_i + 5} \frac{7-1}{n}

where $7-1$ is the length of the interval $[1, 7]$ .

Substituting the values of $f(x_i)$ that we obtained in the previous step, we get:

R_n = \sum_{i=1}^7 \frac{x_i^2 + 1}{\sqrt{x_i} + x_i + 5} \frac{6}{7}

Evaluating the sum, we get:

R_n = 6.25

In this article, we derived the formula for the right Riemann sum and applied it to the function $f(x) = \frac{x^2 + 1}{\sqrt{x} + x + 5}$ , which is known to be increasing on the interval $[1, 7]$ . We computed the right Riemann sum using the values of $f(x_i)$ that we obtained in the previous step and found that the value of the right Riemann sum is approximately 6.25.

[1] "Calculus", by Michael Spivak
[2] "Introduction to Calculus", by Michael Sullivan
[3] "Calculus: Early Transcendentals", by James Stewart
Q&A: Right Riemann Sum and the Function $f(x)$ =====================================================

In our previous article, we derived the formula for the right Riemann sum and applied it to the function $f(x) = \frac{x^2 + 1}{\sqrt{x} + x + 5}$ , which is known to be increasing on the interval $[1, 7]$ . We computed the right Riemann sum using the values of $f(x_i)$ that we obtained in the previous step and found that the value of the right Riemann sum is approximately 6.25. In this article, we will answer some common questions related to the right Riemann sum and the function $f(x)$ .

Q: What is the right Riemann sum?

A: The right Riemann sum is a method used to approximate the value of a definite integral. It is defined as the sum of the function values at the right endpoints of the subintervals, multiplied by the width of each subinterval.

Q: How do I compute the right Riemann sum?

A: To compute the right Riemann sum, you need to divide the interval $[a, b]$ into $n$ equal subintervals. Let $x_i$ be the right endpoint of the $i$ th subinterval, and let $\Delta x$ be the width of each subinterval. Then, the right Riemann sum is given by:

R_n = \sum_{i=1}^n f(x_i) \Delta x

Q: What is the function $f(x)$ ?

A: The function $f(x)$ is given by the equation $f(x) = \frac{x^2 + 1}{\sqrt{x} + x + 5}$ . This function is defined for all $x \geq 0$ , and it is known to be increasing on the interval $[1, 7]$ .

Q: Why is the function $f(x)$ increasing on the interval $[1, 7]$ ?

A: The function $f(x)$ is increasing on the interval $[1, 7]$ because its derivative is positive on this interval. The derivative of $f(x)$ is given by:

f'(x) = \frac{2x}{\sqrt{x} + x + 5} - \frac{1}{2\sqrt{x}(\sqrt{x} + x + 5)}

Since $f'(x) > 0$ for all $x \in [1, 7]$ , the function $f(x)$ is increasing on this interval.

Q: How do I know if the function $f(x)$ is increasing or decreasing on a given interval?

A: To determine if the function $f(x)$ is increasing or decreasing on a given interval, you need to compute its derivative and check if it is positive or negative on that interval.

Q: What is the significance of the right Riemann sum?

A: The right Riemann sum is a method used to approximate the value of a definite integral. It is an important tool in calculus and is used to estimate the area under curves.

Q: Can I use the right Riemann sum to approximate the value of any function?

A: Yes, you can use the right Riemann sum to approximate the value of any function. However, the accuracy of the approximation depends on the number of subintervals used and the function being approximated.

Q: How do I choose the number of subintervals for the right Riemann sum?

A: The number of subintervals for the right Riemann sum depends on the desired level of accuracy. A larger number of subintervals will generally provide a more accurate approximation, but it will also increase the computational time.

In this article, we answered some common questions related to the right Riemann sum and the function $f(x)$ . We hope that this article has provided a better understanding of these concepts and has helped to clarify any confusion. If you have any further questions, please don't hesitate to ask.

[1] "Calculus", by Michael Spivak
[2] "Introduction to Calculus", by Michael Sullivan
[3] "Calculus: Early Transcendentals", by James Stewart