/*! 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 16 Find the indicated Trapezoid Rul... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the indicated Trapezoid Rule approximations to the following integrals. \(\int_{1}^{9} x^{3} d x\) using \(n=2,4,\) and 8 subintervals

Short Answer

Expert verified
The trapezoid rule approximations of the integral of \(x^3\) from 1 to 9 using 2, 4, and 8 subintervals are 2930, 1614, and 1219, respectively.

Step by step solution

01

Set up the integral and identify the function and limits of integration

We are given the integral \(\displaystyle\int_{1}^{9} x^{3} d x\). Here, the function we are integrating is \(f(x)=x^3\) and the limits of integration are \(a=1\) and \(b=9\).
02

Identify the number of subintervals and calculate the width of each subinterval

We need to find approximations using \(n=2\), \(n=4\), and \(n=8\) subintervals. The width of each subinterval, \(\Delta x\), can be calculated as: \(\Delta x = \frac{b-a}{n}\) For \(n=2\), \(\Delta x = \frac{9-1}{2} = 4\) For \(n=4\), \(\Delta x = \frac{9-1}{4} = 2\) For \(n=8\), \(\Delta x = \frac{9-1}{8} = 1\)
03

Find the trapezoid rule approximation for each value of n

Using the trapezoid rule formula given above, we can find the approximations for each value of n: For \(n=2\): \(\displaystyle\int_{1}^{9} x^{3} d x\approx\frac{4}{2}\left[f(1)+2 f(1+4)+f(9)\right]=2[1^3+2(5^3)+9^3]=2[1+2(125)+729]=2930\) For \(n=4\): \(\displaystyle\int_{1}^{9} x^{3} d x\approx\frac{2}{2}\left[f(1)+2 f(1+2)+2 f(1+2+2)+f(9)\right]=1[1^3+2(3^3)+2(5^3)+9^3]=1[1+2(27)+2(125)+729]=1614\) For \(n=8\): \(\displaystyle\int_{1}^{9} x^{3} d x\approx\frac{1}{2}\left[f(1)+2 \sum_{i=1}^{7} f(1+i)+f(9)\right]=\frac{1}{2}\left[1+2[2^3+3^3+4^3+5^3+6^3+7^3+8^3]+9^3\right]=\frac{1}{2}(1+2[1806]+729)=1219\) So, the trapezoid rule approximations of the given integral using \(n=2\), \(n=4\), and \(n=8\) subintervals are 2930, 1614, and 1219, respectively.

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Key Concepts

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

Numerical Integration
Numerical integration is a useful technique for finding the approximate value of integrals, especially when an analytical solution is difficult or impossible to obtain. In mathematics, integration is a fundamental operation used to calculate areas under curves or to aggregate values continuously. However, calculating integrals directly can sometimes be complex. This is where numerical integration comes in handy.

Instead of finding an exact solution, numerical methods, like the Trapezoid Rule, allow us to estimate the value of functions over an interval. These methods split the area into smaller segments and calculate the aggregate area of these segments to approximate the integral.
  • The Trapezoid Rule is one of the simplest methods, relying on using trapezoids rather than rectangles to approximate the area under a curve.
  • By increasing the number of trapezoids, or subintervals, the approximation gets closer to the true integral value.
Numerical integration plays a vital role in fields ranging from physics and engineering to finance and beyond.
Integral Approximations
Integral approximations are methods used to estimate the value of integrals when exact computation is not feasible. The Trapezoid Rule is an example of such an approximation technique. In this method, the curve is divided into several trapezoids, and their areas are summed to estimate the integral.

The formula for the Trapezoid Rule is given as:\[\int_{a}^{b} f(x) \, dx \approx \frac{\Delta x}{2} \left[f(x_0) + 2 \sum_{i=1}^{n-1} f(x_i) + f(x_n)\right]\]
  • Here, \(\Delta x\) is the width of each subinterval.
  • \(x_0, x_1, \ldots, x_n\) are points that divide the interval \([a, b]\) into \(n\) subintervals.
As the number of subintervals increases, the trapezoids fit the curve more closely, improving the approximation. The balance between computational cost and desired accuracy is crucial when choosing the number of subintervals.
Subintervals
Subintervals are divisions of the integration range \([a, b]\) into smaller sections, enhancing the accuracy of numerical approximations. When applying the Trapezoid Rule, the entire interval is split into equal parts, known as subintervals, each having the width \(\Delta x\).

For example, in the integral \(\int_{1}^{9} x^{3} \, dx\), we used subintervals with \(n=2, 4,\) and \(8\). The calculation of \(\Delta x\) for each is:
  • For \(n=2\), \(\Delta x = \frac{9-1}{2} = 4\).
  • For \(n=4\), \(\Delta x = \frac{9-1}{4} = 2\).
  • For \(n=8\), \(\Delta x = \frac{9-1}{8} = 1\).
More subintervals mean narrower widths, allowing better conformity of the trapezoids to the curve. This results in a more precise approximation of the integral's value. Finding the optimal number of subintervals is key in balancing accuracy and computational efficiency.

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Most popular questions from this chapter

Compare the errors in the Midpoint and Trapezoid Rules with \(n=4,8,16,\) and 32 subintervals when they are applied to the following integrals (with their exact values given). \(\int_{0}^{\pi} \ln (5+3 \cos x) d x=\pi \ln \frac{9}{2}\)

Consider the general first-order initial value problem \(y^{\prime}(t)=a y+b, y(0)=y_{0},\) for \(t \geq 0,\) where \(a, b,\) and \(y_{0}\) are real numbers. a. Explain why \(y=-b / a\) is an equilibrium solution and corresponds to horizontal line segments in the direction field. b. Draw a representative direction field in the case that \(a>0\). Show that if \(y_{0}>-b / a,\) then the solution increases for \(t \geq 0\) and if \(y_{0}<-b / a,\) then the solution decreases for \(t \geq 0\). c. Draw a representative direction field in the case that \(a<0\). Show that if \(y_{0}>-b / a,\) then the solution decreases for \(t \geq 0\) and if \(y_{0}<-b / a,\) then the solution increases for \(t \geq 0\).

Use the window \([-2,2] \times[-2,2]\) to sketch a direction field for the following equations. Then sketch the solution curve that corresponds to the given initial condition. $$y^{\prime}(t)=\sin y, y(-2)=\frac{1}{2}$$

A remarkable integral It is a fact that \(\int_{0}^{\pi / 2} \frac{d x}{1+\tan ^{m} x}=\frac{\pi}{4}\) for all real numbers \(m .\) a. Graph the integrand for \(m=-2,-3 / 2,-1,-1 / 2,0,1 / 2\) \(1,3 / 2,\) and \(2,\) and explain geometrically how the area under the curve on the interval \([0, \pi / 2]\) remains constant as \(m\) varies. b. Use a computer algebra system to confirm that the integral is constant for all \(m.\)

\(A n\) integrand with trigonometric functions in the numerator and denominator can often be converted to a rational integrand using the substitution \(u=\tan (x / 2)\) or equivalently \(x=2 \tan ^{-1} u .\) The following relations are used in making this change of variables. \(A: d x=\frac{2}{1+u^{2}} d u \quad B: \sin x=\frac{2 u}{1+u^{2}} \quad C: \cos x=\frac{1-u^{2}}{1+u^{2}}\) $$\text { Evaluate } \int_{0}^{\pi / 3} \frac{\sin \theta}{1-\sin \theta} d \theta$$

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