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Chapter 5: Problem 7

Use the given values of \(a\) and \(b\) to express the following limits as integrals. (Do not evaluate the integrals.) $$\lim _{\max \Delta x_{k} \rightarrow 0} \sum_{k=1}^{n} 4 x_{k}^{*}\left(1-3 x_{k}^{*}\right) \Delta x_{k} ; a=-3, b=3$$

Short Answer

Expert verified
The limit is expressed as the integral \( \int_{-3}^{3} 4x(1-3x) \, dx \).

Step by step solution

01

Identify the Riemann Sum Format

The given expression \( \lim _{\max \Delta x_{k} \rightarrow 0} \sum_{k=1}^{n} 4 x_{k}^{*}(1-3x_{k}^{*}) \Delta x_{k} \) is a Riemann sum. The general form of a Riemann sum is \( \sum_{k=1}^{n} f(x_k^*) \Delta x_k \), which approximates the integral of a function \( f(x) \). Here, \( f(x) = 4x(1-3x) \), \( a = -3 \), and \( b = 3 \).
02

Recognize Integration Limits and Function

The Riemann sum has integration limits from \( a = -3 \) to \( b = 3 \). The function to integrate is \( f(x) = 4x(1-3x) \), which is the expression inside the summation \( \sum_{k=1}^{n} 4 x_{k}^{*}(1-3x_{k}^{*}) \Delta x_{k} \).
03

Express Limit as an Integral

The limit of the Riemann sum as \( \max \Delta x_{k} \to 0 \) represents the definite integral of the function over the interval from \( a \) to \( b \). Therefore, the Riemann sum \( \lim _{\max \Delta x_{k} \rightarrow 0} \sum_{k=1}^{n} 4 x_{k}^{*}(1-3x_{k}^{*}) \Delta x_{k} \) can be expressed as the integral \[ \int_{-3}^{3} 4x(1-3x) \, dx \].

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Key Concepts

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

Definite Integral
In calculus, a definite integral is a way to calculate the area under a curve between two specified points on the x-axis. The notation for a definite integral is \( \int_{a}^{b} f(x) \, dx \), where \( f(x) \) is the function you are integrating, and \( a \) and \( b \) are the lower and upper limits of the integration, respectively. These limits indicate the interval over which we are calculating the area.

  • The definite integral computes the total accumulation of quantities, like area, volume or other quantities, over a given interval.
  • The process essentially adds up infinite, infinitesimally small quantities, which is possible through the limit concept.
Calculating a definite integral involves finding the antiderivative of the function (if possible) and then applying the Fundamental Theorem of Calculus, which states that if \( F(x) \) is an antiderivative of \( f(x) \), then:

\[\int_{a}^{b} f(x) \, dx = F(b) - F(a).\]
Integration
Integration is one of the core operations in calculus, alongside differentiation. It is essentially the process of finding the integral, or in geometric terms, the area under a curve. There are two main types of integration: definite and indefinite.

  • **Definite integrals** calculate the exact area under a curve between two points \( a \) and \( b \), as discussed earlier.
  • **Indefinite integrals**, on the other hand, find the general form of the antiderivative, providing a family of functions \( F(x) + C \), where \( C \) is the constant of integration.
Integration can be visualized as the reverse process of differentiation, which is why it is often referred to as antidifferentiation. Techniques for performing integration include:

  • **Substitution**, which is useful when an integral involves a composition of functions.
  • **Integration by parts**, a method based on the product rule for differentiation, handy for integrals of products of functions.
Calculus
Calculus is a branch of mathematics focused on the study of change and motion. It's divided into two main areas: differential calculus and integral calculus. Each deals with limiting processes that analyze small-scale changes.

  • **Differential calculus** concerns itself with finding the rate at which things change. It uses derivatives to study the slope of curves and the rates of change.
  • **Integral calculus**, on the other hand, deals with accumulation and quantifying areas under and between curves. It uses integrals.
These two areas of calculus are fundamentally linked by the Fundamental Theorem of Calculus, which bridges the concept of differentiation and integration, showing that they are inverse processes.

The applications of calculus are vast and include everything from physics, where it helps predict trajectories in space, to economics, where it models how supply and demand change. It also plays a critical role in fields like engineering and biology.

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Most popular questions from this chapter

Sketch the region described and find its area. The region under the curve \(y=x^{2}+1\) and over the interval [0,3]

An artist wants to create a rough triangular design using uniform square tiles glued edge to edge. She places \(n\) tiles in a row to form the base of the triangle and then makes each successive row two tiles shorter than the preceding row. Find a formula for the number of tiles used in the design. [Hint: Your answer will depend on whether \(n\) is even or odd. \(]\)

(a) Use a graphing utility to generate the graph of $$ f(x)=\frac{1}{100}(x+2)(x+1)(x-3)(x-5) $$ and use the graph to make a conjecture about the sign of the integral \(\int_{-2}^{5} f(x) d x\) (b) Check your conjecture by evaluating the integral.

Use a calculating utility to find the midpoint approximation of the integral using \(n=20\) sub-intervals, and then find the exact value of the integral using Part 1 of the Fundamental Theorem of Calculus. $$\int_{0}^{\pi / 2} \sin x d x$$

$$\begin{aligned} \sum_{k=1}^{4}\left[(k+1)^{3}-k^{3}\right]=&\left[5^{3}-4^{3}\right]+\left[4^{3}-3^{3}\right] \\\ &+\left[3^{3}-2^{3}\right]+\left[2^{3}-1^{3}\right] \\ =& 5^{3}-1^{3}=124 \end{aligned}$$ 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. $$\sum_{k=2}^{20}\left(\frac{1}{k^{2}}-\frac{1}{(k-1)^{2}}\right)$$
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