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If the Midpoint Rule is used on the interval [-1,11] with \(n=3\) subintervals, at what \(x\) -coordinates is the integrand evaluated?

Short Answer

Expert verified
Answer: The integrand is evaluated at the x-coordinates x = 1, x = 5, and x = 9.

Step by step solution

01

Determine the width of each subinterval

To find the width of each subinterval, we can use the formula \(\Delta x = \frac{b-a}{n}\), where \(b\) is the right endpoint of the interval, \(a\) is the left endpoint, and \(n\) is the number of subintervals. In this case, \(a = -1\), \(b = 11\), and \(n = 3\). Plugging these values into the formula, we get: \(\Delta x = \frac{11-(-1)}{3} = \frac{12}{3} = 4\) So the width of each subinterval is 4.
02

Find the midpoints of each subinterval

With a width of 4 for each subinterval, the interval [-1, 11] can be divided into 3 subintervals: [-1, 3], [3, 7], and [7, 11]. The midpoint of each subinterval is the average of its endpoints: Midpoint of [-1,3] = \(\frac{(-1) + 3}{2} = 1\) Midpoint of [3,7] = \(\frac{3 + 7}{2} = 5\) Midpoint of [7,11] = \(\frac{7 + 11}{2} = 9\)
03

List the \(x\)-coordinates at which the integrand is evaluated

Using the Midpoint Rule, we evaluate the integrand at the midpoints of each subinterval. In this case, the midpoints are 1, 5, and 9. So, the integrand is evaluated at the \(x\)-coordinates \(x=1\), \(x=5\), and \(x=9\).

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Key Concepts

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

Numerical Integration
When dealing with complex equations or functions that are not easily integrable by standard analytical methods, numerical integration becomes a helpful tool. It provides an approximate solution by breaking down the integral into more manageable parts. One popular numerical method is the Midpoint Rule. This method estimates the integral of a function over an interval by dividing it into smaller subintervals and evaluating the function at the midpoint of each subinterval. This approach is particularly useful in cases where precision is important, yet traditional methods would be too cumbersome.

Numerical integration methods like the Midpoint Rule prove invaluable in fields such as physics, engineering, and computer science, where exact integrals are often too complex or impossible to find analytically. Using numerical calculations, we can approximate these complex areas under curves effectively, ensuring practical applications across various scientific disciplines.

  • Approximate area under curves.
  • Useful when the integrand is complex.
  • Applies to multiple scientific disciplines.
Subintervals
In the context of numerical integration, dividing an interval into smaller sections called subintervals is a key step. The number of subintervals, denoted by \(n\), dictates the accuracy of the approximation. More subintervals typically mean a more accurate estimate, as each portion of the curve is taken into account in finer detail. In the exercise at hand, the interval \([-1, 11]\) was divided into 3 subintervals, each with a width calculated using the formula \(\Delta x = \frac{b-a}{n}\).

Determining the range and size of each subinterval helps in focusing the evaluation precisely where needed. Midpoints or specific function values calculated within these subintervals allow for applying numerical methods like the Midpoint Rule to approximate integrals effectively. Factors like computational efficiency and desired precision determine the number of subintervals used.

  • Increases accuracy of numerical approximations.
  • Determines the points of evaluation within an interval.
  • Balances between precision and computational resources.
Calculus
Calculus provides the fundamental techniques for understanding change and is crucial for analyzing the realities of the natural world. Integral calculus, a major component of calculus, deals with finding areas under curves, which is essentially what integration aims to achieve. When small changes are taken over an interval, we use integration to sum them up, providing us with a holistic view of a function's behavior over that range.

In the exercise provided, calculus is applied through numerical methods to approximate these areas under a curve—integrating without directly computing the exact integral. While traditional calculus allows us to compute integrals using antiderivatives or standard techniques, numerical methods like the Midpoint Rule offer practical solutions when dealing with more complex or computationally demanding problems. This blend of calculus with numerical methods shows the adaptability of mathematical concepts to different types of problems.

  • Integral calculus focuses on summing changes over intervals.
  • Useful for analyzing real-world phenomena.
  • Supports numerical methods in complex situations.

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Most popular questions from this chapter

Many methods needed Show that \(\int_{0}^{\infty} \frac{\sqrt{x} \ln x}{(1+x)^{2}} d x=\pi\) in the following steps. a. Integrate by parts with \(u=\sqrt{x} \ln x.\) b. Change variables by letting \(y=1 / x.\) c. Show that \(\int_{0}^{1} \frac{\ln x}{\sqrt{x}(1+x)} d x=-\int_{1}^{\infty} \frac{\ln x}{\sqrt{x}(1+x)} d x\) and conclude that \(\int_{0}^{\infty} \frac{\ln x}{\sqrt{x}(1+x)} d x=0.\) d. Evaluate the remaining integral using the change of variables \(z=\sqrt{x}\) (Source: Mathematics Magazine 59, No. 1 (February 1986): 49).

\(A\) powerful tool in solving problems in engineering and physics is the Laplace transform. Given a function \(f(t),\) the Laplace transform is a new function \(F(s)\) defined by $$ F(s)=\int_{0}^{\infty} e^{-s t} f(t) d t $$ where we assume that s is a positive real number. For example, to find the Laplace transform of \(f(t)=e^{-t},\) the following improper integral is evaluated: $$ F(s)=\int_{0}^{\infty} e^{-s t} e^{-t} d t=\int_{0}^{\infty} e^{-(s+1) t} d t=\frac{1}{s+1} $$ Verify the following Laplace transforms, where a is a real number. $$f(t)=\cos a t \longrightarrow F(s)=\frac{s}{s^{2}+a^{2}}$$

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Use numerical methods or a calculator to approximate the following integrals as closely as possible. $$\int_{0}^{\infty} \frac{\sin ^{2} x}{x^{2}} d x=\frac{\pi}{2}$$

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