/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 4 Find the value of (a) \(\sum_{k=... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the value of (a) \(\sum_{k=1}^{n} f\left(x_{k}^{*}\right) \Delta x_{k}\) (b) \(\max \Delta x_{k}\). $$\begin{aligned} &f(x)=x^{3} ; a=-3, b=3 ; n=4;\\\ &\Delta x_{1}=2, \Delta x_{2}=1, \Delta x_{3}=1, \Delta x_{4}=2;\\\ &x_{1}^{*}=-2, x_{2}^{*}=0, x_{3}^{*}=0, x_{4}^{*}=2 \end{aligned}$$

Short Answer

Expert verified
(a) 0; (b) 2.

Step by step solution

01

Calculate Function Values

Substitute each sample point value, \(x^*_k\), into the function \(f(x) = x^3\) to get \(f(x^*_k)\). We have the following calculations:- \(f(x_1^*) = f(-2) = (-2)^3 = -8\)- \(f(x_2^*) = f(0) = (0)^3 = 0\)- \(f(x_3^*) = f(0) = (0)^3 = 0\)- \(f(x_4^*) = f(2) = (2)^3 = 8\)
02

Compute Each Term in the Series

Multiply each \(f(x^*_k)\) by its corresponding \(\Delta x_k\) to find each product:- \(f(x_1^*) \Delta x_1 = -8 \times 2 = -16\)- \(f(x_2^*) \Delta x_2 = 0 \times 1 = 0\)- \(f(x_3^*) \Delta x_3 = 0 \times 1 = 0\)- \(f(x_4^*) \Delta x_4 = 8 \times 2 = 16\)
03

Sum the Series

Add up all the products calculated in Step 2 to find the sum of the series:\(\sum_{k=1}^{n} f(x_k^*) \Delta x_k = -16 + 0 + 0 + 16 = 0\)
04

Find Maximum \( \Delta x_k \)

Identify the largest subinterval length \(\Delta x_k\) from the given intervals:- Comparing \(\Delta x_1 = 2\), \(\Delta x_2 = 1\), \(\Delta x_3 = 1\), \(\Delta x_4 = 2\), we see that the largest is \(\Delta x_1 = 2\) and \(\Delta x_4 = 2\).Thus, \(\max \Delta x_k = 2\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Definite Integral
A definite integral is a powerful concept in calculus that allows us to calculate the total accumulation of a function over an interval. Imagine you want to find the total area under a curve and above the x-axis, within a certain range of x values. This is where the definite integral comes into play. It is represented by the notation \( \int_{a}^{b} f(x) \, dx \), where \( a \) is the starting point and \( b \) is the endpoint of the interval.
### Why Use a Definite Integral?- **Total Accumulation**: It helps in finding quantities like total distance, area, and volume.- **Summation Approach**: Think of it as adding up infinitely small pieces to get a whole.- **Limits**: Provides precise calculations by taking the limits of sums as the pieces get infinitely small.
When dealing with integrals, the limits \( a \) and \( b \) are crucial as they define the boundary over which you want to accumulate the values of the function \( f(x) \). Instead of manually summing up the quantity of interest, definite integrals make these tasks more achievable, mathematically sound, and less error-prone.
Finite Subdivision
To find a definite integral using the Riemann Sum method, you first need to divide the interval [a, b] into smaller chunks. This process is known as finite subdivision. By splitting the interval, you prepare it for further analysis by creating several subintervals.
When you divide the interval into these smaller segments, each subinterval has a width \( \Delta x_k \). The sum of these segments approximates the area under the curve. Let’s say your interval is from \( a \) to \( b \) and you want to divide it into \( n \) equal parts. You will get \( n \) subintervals of width \( \Delta x = \frac{b-a}{n} \). This divides your main problem into smaller tasks.
### Key Points of Finite Subdivision- **Flexibility**: Not all subintervals need be equal.- **Precision**: More subintervals lead to a more precise approximation.- **Visualize**: Think of these as pieces of a puzzle that together form the complete picture.
In the original exercise, we used four subintervals of different sizes, which allowed us to gain insight into variations in slope or other aspects of the function's curve.
Sample Points
Sample points are key to understanding and deploying Riemann Sums effectively. Imagine them as reference markers within each subinterval. In a given subinterval \( [x_{k-1}, x_k] \), the sample point \( x_k^* \) is a chosen point at which the function \( f(x) \) will be evaluated. This evaluation helps in estimating the contribution of each subinterval to the total sum.
### How to Select Sample Points- **Midpoint Method**: Choose the midpoint for better approximation.- **Endpoint Method**: Use either the left or right endpoint; simpler but less accurate.- **Random Point**: Sometimes randomly chosen for particular types of analysis.
In the solution provided, you can see sample points \( x_1^* = -2, x_2^* = 0, x_3^* = 0, x_4^* = 2 \) used to calculate \( f(x_k^*) \). These are arbitrarily chosen within the given subintervals. The choice depends on the accuracy needed and the nature of the problem you are solving. The function is evaluated at these points, and when multiplied by \( \Delta x_k \), contributes to the overall Riemann Sum, giving you the approximation you need.

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Most popular questions from this chapter

Evaluate each integral by first modifying the form of the integrand and then making an appropriate substitution, if needed. $$\int \frac{t+1}{t} d t$$

(a) Prove that if \(f\) is an odd function, then $$\int_{-a}^{a} f(x) d x=0$$ and give a geometric explanation of this result. [Hint: One way to prove that a quantity \(q\) is zero is to show that \(q=-q .]\) (b) Prove that if \(f\) is an even function, then $$\int_{-a}^{a} f(x) d x=2 \int_{0}^{a} f(x) d x$$ and give a geometric explanation of this result. [Hint: Split the interval of integration from \(-a\) to \(a\) into two parts at 0.]

$$\begin{aligned} \sum_{k=1}^{4}\left[(k+1)^{3}-k^{3}\right]=&\left[5^{3}-4^{3}\right]+\left[4^{3}-3^{3}\right] \\\ &+\left[3^{3}-2^{3}\right]+\left[2^{3}-1^{3}\right] \\ =& 5^{3}-1^{3}=124 \end{aligned}$$ For convenience, the terms are listed in reverse order. Note how cancellation allows the entire sum to collapse like a telescope. A sum is said to telescope when part of each term cancels part of an adjacent term, leaving only portions of the first and last terms uncanceled. Evaluate the telescoping sums in these exercises. $$\sum_{k=5}^{17}\left(3^{k}-3^{k-1}\right)$$

(a) If \(h^{\prime}(t)\) is the rate of change of a child's height measured in inches per year, what does the integral \(\int_{0}^{10} h^{\prime}(t) d t\) represent, and what are its units? (b) If \(r^{\prime}(t)\) is the rate of change of the radius of a spherical balloon measured in centimeters per second, what does the integral \(\int_{1}^{2} r^{\prime}(t) d t\) represent, and what are its units? (c) If \(H(t)\) is the rate of change of the speed of sound with respect to temperature measured in \(\mathrm{ft} / \mathrm{s}\) per "F. what does the integral \(\int_{32}^{100} H(t) d t\) represent. and what are its units? (d) If \(v(t)\) is the velocity of a particle in rectilinear motion, measured in \(\mathrm{cm} / \mathrm{h}\), what does the integral \(\int_{t_{1}}^{t_{2}} v(t) d t\) represent, and what are its units?

Evaluate (a) \(\int_{-1}^{1} x \sqrt{\cos \left(x^{2}\right)} d x\). (b) \(\int_{0}^{\pi} \sin ^{8} x \cos ^{5} x d x\). [Hint: Use the substitution \(u=x-(\pi / 2) .\)]

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