### Numerical Estimation in Calculus**Problem Statement:**1. Use a calculator to give a ballpark estimate of each of $ S_{2^{16}} $, $ S_{2^{24}} $, and $ S_{2^{32}} $. The required degree of accuracy, in terms of the number of digits, is up to the shown number of boxes.#### Estimation Boxes:- For $ S_{2^{16}} $: - $\begin{cases} 6 & \boxed{} \boxed{} \boxed{} \boxed{} \boxed{} \boxed{} \boxed{} \boxed{} \boxed{} \dots & \text{(use left-end numbers)} \\ 6 & \boxed{} \boxed{} \boxed{} \boxed{} \boxed{} \boxed{} \boxed{} \boxed{} \boxed{} \dots & \text{(use right-end numbers)} \\ \end{cases} $- For $ S_{2^{24}} $: - $\begin{cases} 6 & \boxed{} \boxed{} \boxed{} \boxed{} \boxed{} \boxed{} \boxed{} \boxed{} \boxed{} \dots & \text{(use left-end numbers)} \\ 6 & \boxed{} \boxed{} \boxed{} \boxed{} \boxed{} \boxed{} \boxed{} \boxed{} \boxed{} \dots & \text{(use right-end numbers)} \\ \end{cases} $- For $ S_{2^{32}} $: - $\begin{cases} 6 & \boxed{} \boxed{} \boxed{} \boxed{} \boxed{} \boxed{} \boxed{} \boxed{} \boxed{} \dots & \text{(use left-end numbers)} \\ 6 & \boxed{} \boxed{} \boxed{} \boxed{} \boxed{} \boxed{} \boxed{} \boxed{} \boxed{} \dots & \text{(use right-end numbers)} \\ \end{cases} $2. Based on the formula for $ S_{2^t} $ given on the previous page, conclude:\[\int_{x=-2}^{4} x^5 \, dx = \frac{1}{6} \left( \boxed{} \right) - \cdots\ ### Riemann Sums with Reference Function**Function:**\[ f(x) = x^5 \]**Interval:**\[ \left[ -2, 4 \right] \]**Subintervals:**Using $2^\ell$ subintervals of equal length.**Riemann Sum:**The Riemann sum is generally used to approximate the integral of a function over an interval. In this case, the Riemann sum using the reference function $f(x) = x^5$ over the interval $\left[ -2, 4 \right]$, using $2^\ell$ subintervals of equal length, is given by: \[ S_{2^\ell} = \begin{cases} 48 \cdot \left(14 - 66 \cdot \left( \frac{1}{2^\ell} \right) + 75 \cdot \left( \frac{1}{2^\ell} \right)^2 - 27 \cdot \left( \frac{1}{2^\ell} \right)^4\right) & \text{(use left-end numbers)}, \\ 48 \cdot \left(14 + 66 \cdot \left( \frac{1}{2^\ell} \right) + 75 \cdot \left( \frac{1}{2^\ell} \right)^2 - 27 \cdot \left( \frac{1}{2^\ell} \right)^4\right) & \text{(use right-end numbers)}. \end{cases} \]**Explanation of Notation:**- **$48 \cdot \left(\cdots\right)$**: This factor is likely derived from the length of the interval and constants of the Riemann sum formula.- **Left-end numbers**: Use the left endpoints of each subinterval to calculate the sum.- **Right-end numbers**: Use the right endpoints of each subinterval to calculate the sum.**Components of the Sum:**- **$14$**: Constant term.- **$-66 \cdot \left( \frac{1}{2^\ell} \right)$** or **$66 \cdot \left( \frac{1}{2^\ell} \right)$**: First-order term

Consider the provided question,

The Riemann sum, with the reference function (ad f (a) over the interval- 2, 4, using 2 subintervals of equal length, is given by (G) + 76 ()- (+)') 2 1 (+) 48 - 14 – 66- + 75 - - 27· 22 - 2e 24 slue use left-end umbers), Sze () ()') 4 48 - 14 + 66· + 75- 26 - 27· - 2l use right-end numbers ).

The Riemann sum, with the reference function (ad f (a) over the interval- 2, 4, using 2 subintervals of equal length, is given by (G) + 76 ()- (+)') 2 1 (+) 48 - 14 – 66- + 75 - - 27· 22 - 2e 24 slue use left-end umbers), Sze () ()') 4 48 - 14 + 66· + 75- 26 - 27· - 2l use right-end numbers ).

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Concept explainers

Riemann Sum

Riemann Sums is a special type of approximation of the area under a curve by dividing it into multiple simple shapes like rectangles or trapezoids and is used in integrals when finite sums are involved. Figuring out the area of a curve is complex hence this method makes it simple. Usually, we take the help of different integration methods for this purpose. This is one of the major parts of integral calculus.

Riemann Integral

Bernhard Riemann's integral was the first systematic description of the integral of a function on an interval in the branch of mathematics known as real analysis.

Question

### Numerical Estimation in Calculus

**Problem Statement:**

1. Use a calculator to give a ballpark estimate of each of $ S_{2^{16}} $, $ S_{2^{24}} $, and $ S_{2^{32}} $.
The required degree of accuracy, in terms of the number of digits, is up to the shown number of boxes.

#### Estimation Boxes:

- For $ S_{2^{16}} $:
- $\begin{cases}
6 & \boxed{} \boxed{} \boxed{} \boxed{} \boxed{} \boxed{} \boxed{} \boxed{} \boxed{} \dots & \text{(use left-end numbers)} \\
6 & \boxed{} \boxed{} \boxed{} \boxed{} \boxed{} \boxed{} \boxed{} \boxed{} \boxed{} \dots & \text{(use right-end numbers)} \\
\end{cases}
$

- For $ S_{2^{24}} $:
- $\begin{cases}
6 & \boxed{} \boxed{} \boxed{} \boxed{} \boxed{} \boxed{} \boxed{} \boxed{} \boxed{} \dots & \text{(use left-end numbers)} \\
6 & \boxed{} \boxed{} \boxed{} \boxed{} \boxed{} \boxed{} \boxed{} \boxed{} \boxed{} \dots & \text{(use right-end numbers)} \\
\end{cases}
$

- For $ S_{2^{32}} $:
- $\begin{cases}
6 & \boxed{} \boxed{} \boxed{} \boxed{} \boxed{} \boxed{} \boxed{} \boxed{} \boxed{} \dots & \text{(use left-end numbers)} \\
6 & \boxed{} \boxed{} \boxed{} \boxed{} \boxed{} \boxed{} \boxed{} \boxed{} \boxed{} \dots & \text{(use right-end numbers)} \\
\end{cases}
$

2. Based on the formula for $ S_{2^t} $ given on the previous page, conclude:

\[\int_{x=-2}^{4} x^5 \, dx = \frac{1}{6} \left( \boxed{} \right) - \cdots\

### Riemann Sums with Reference Function

**Function:**

\[ f(x) = x^5 \]

**Interval:**

\[ \left[ -2, 4 \right] \]

**Subintervals:**

Using $2^\ell$ subintervals of equal length.

**Riemann Sum:**

The Riemann sum is generally used to approximate the integral of a function over an interval. In this case, the Riemann sum using the reference function $f(x) = x^5$ over the interval $\left[ -2, 4 \right]$, using $2^\ell$ subintervals of equal length, is given by:

\[
S_{2^\ell} =
\begin{cases}
48 \cdot \left(14 - 66 \cdot \left( \frac{1}{2^\ell} \right) + 75 \cdot \left( \frac{1}{2^\ell} \right)^2 - 27 \cdot \left( \frac{1}{2^\ell} \right)^4\right) & \text{(use left-end numbers)}, \\

48 \cdot \left(14 + 66 \cdot \left( \frac{1}{2^\ell} \right) + 75 \cdot \left( \frac{1}{2^\ell} \right)^2 - 27 \cdot \left( \frac{1}{2^\ell} \right)^4\right) & \text{(use right-end numbers)}.
\end{cases}
\]

**Explanation of Notation:**

- **$48 \cdot \left(\cdots\right)$**: This factor is likely derived from the length of the interval and constants of the Riemann sum formula.
- **Left-end numbers**: Use the left endpoints of each subinterval to calculate the sum.
- **Right-end numbers**: Use the right endpoints of each subinterval to calculate the sum.

**Components of the Sum:**

- **$14$**: Constant term.
- **$-66 \cdot \left( \frac{1}{2^\ell} \right)$** or **$66 \cdot \left( \frac{1}{2^\ell} \right)$**: First-order term

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