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Chapter 5: Problem 17

\(15-18=\) Express the limit as a definite integral on the given interval. $$\lim _{n \rightarrow \infty} \sum_{i=1}^{n}\left[5\left(x_{i}^{*}\right)^{3}-4 x_{i}^{*}\right] \Delta x, \quad[2,7]$$

Short Answer

Expert verified
\(\int_{2}^{7} (5x^3 - 4x) \, dx\)

Step by step solution

01

Identify the function and interval

In the given expression, \[ \lim _{n \rightarrow \infty} \sum_{i=1}^{n}\left[5\left(x_{i}^{*}\right)^{3}-4 x_{i}^{*}\right] \Delta x \] the function inside the summation is \( f(x) = 5x^3 - 4x \), and the interval is \([2,7]\).
02

Understand the Riemann Sum

The given expression for the limit represents a Riemann sum, which approximates the area under the curve \( f(x) = 5x^3 - 4x \) over the interval \([2,7]\). As \( n \) approaches infinity, the Riemann sum converges to the exact area, which is expressed as a definite integral.
03

Express the Riemann Sum as a Definite Integral

Using the Riemann sum definition, we convert the expression into a definite integral:\[ \int_{2}^{7} (5x^3 - 4x) \, dx \]. This integral represents the exact area under the curve of \( f(x) \) from \( x=2 \) to \( x=7 \).

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Key Concepts

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

Riemann Sum
When dealing with integrals, the Riemann sum is a fundamental concept. It helps us to approximate the area under a curve. Imagine slicing the region under a curve into many thin rectangles. Each rectangle's width is given by \( \Delta x \), and its height is determined by the function value at a specific point within that interval, which is represented as \( x_i^* \).
  • The basic idea is to calculate the sum of the areas of all these rectangles.
  • Mathematically, this is expressed as \( \sum_{i=1}^{n} f(x_i^*) \Delta x \).
  • Here, \( f(x_i^*) \) is the height of each rectangle.
  • As the number of rectangles \( n \) increases, \( \Delta x \), which is the width of each rectangle, becomes very small.
The process improves our approximation of the area under the curve. It is one way to understand how integration captures the essence of area accumulation.
Limit
In calculus, the concept of a limit helps us understand behavior as things change, often when values approach a certain point. Regarding Riemann sums, taking the limit as \( n \to \infty \) means we're increasing the number of rectangles indefinitely.
  • Increasing \( n \) makes each rectangle thinner, giving a finer approximation.
  • The limit essentially captures the idea of making \( \Delta x \) infinitesimally small.
  • When taken to infinity, the sum of the rectangle areas approaches the true area under the curve.
This approach of taking a limit is what transitions a Riemann sum into an integral, offering an exact calculation rather than an approximation.
Function Approximation
At its core, approximating a function involves simplifying complex shapes into more understandable geometric forms, like rectangles. Riemann sums and limits combine to approximate a function's behavior over an interval.
  • For the function \( f(x) = 5x^3 - 4x \), this involves estimating the area under its curve on the interval [2, 7].
  • This approximation becomes very accurate as \( n \) grows, improving the estimate of the integral.
  • As our rectangles amount toward infinity, their collective area sums up to the integral.
Function approximation, especially through the Riemann sum and limit, is key in switching from discrete summations to continuous integrals, allowing us to understand changes and trends in a measurable and complete manner.

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