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Limits and Integrals In Exercises 73 and 74 , evaluate the limit and sketch the graph of the region whose area is represented by the limit. $$\lim _ { \| \Delta \| \rightarrow 0 } \sum _ { i = 1 } ^ { n } \left( x _ { i } - x _ { i } ^ { 2 } \right) \Delta x ,\( where \)x _ { i } = \frac { i } { n }\( and \)\Delta x = \frac { 1 } { n }$$

Short Answer

Expert verified
The area under the curve as represented by the limit is \(\frac{1}{6}\).

Step by step solution

01

Identify the function

Firstly, identify the function from the given expression. Here, the function \(f(x_i)\) represented in the sum is \(x-x^2\). The \(\Delta x\) is \(\frac {1}{n}\) and \(x_i\) is \(\frac {i}{n}\).
02

Express the limit as a definite integral

Next, express the given limit of sum as a definite integral. In general, the limit of a Riemann sum as the norm of the partition approaches zero equals the definite integral. So, \[\lim_{\| \Delta \| \rightarrow 0} \sum_{i = 1}^{n}\left(x_{i} - x_{i}^2\right) \Delta x = \int_0^1 (x-x^2) dx\]
03

Evaluate the definite integral

Now, compute this definite integral using the Fundamental Theorem of Calculus. \[\int_0^1 (x-x^2) dx = \left[\frac{x^2}{2} - \frac{x^3}{3}\right]_0^1 \] This gives us \(\frac{1}{2} - \frac{1}{3} = \frac{1}{6}\)

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Key Concepts

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

Limits in Calculus
Limits are a fundamental part of calculus and essential to understanding the behavior of functions as they approach a specific point. When students encounter the concept of a limit in calculus, it's crucial that they grasp the idea of approaching a value without necessarily reaching it. In the context of Riemann sums and definite integrals, we use limits to transition from summation of finite partitions to the exact area under a curve.

A classic example is the limit definition of a definite integral. You may have seen an expression like \(\lim_{\| \Delta x \| \rightarrow 0} \sum_{i=1}^{n} f(x_i)\Delta x\). This essentially says that as the partitions of the interval over which we're summing become finer and finer (making \(\Delta x\) very small), the sum of the areas of the rectangles approaches the actual area under the function \(f(x)\). In practice, by making the partitions infinitely small, we find the exact area, and that's where the concept of a definite integral comes into play.
Definite Integral
The definite integral is a central idea in calculus representing the net area under a curve between two points. It's defined as the limit of a Riemann sum as the partition of the interval becomes infinitely fine. When a student solves a definite integral, they're essentially finding the accumulated change or total area under the curve over [a, b], the interval of integration.

In the exercise, we deal with the integral \(\int_0^1 (x-x^2) dx\), which can be visualized as the area under the curve \(y = x - x^2\) from \(x = 0\) to \(x = 1\). To ensure the topic is understood, it's essential to reinforce what the integral notation means: \(\int_{a}^{b} f(x) dx\) is the sum of infinitesimally small products of the function value \(f(x)\) and the infinitely small width \(dx\), across the interval \(a\) to \(b\). It's a concept that ties together the infinite and the finite, giving us a powerful tool to measure continuous changes.
Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus (FTC) is a monumental achievement in mathematics that connects differentiation and integration. Understanding the FTC is essential for students, as it provides an efficient method to evaluate definite integrals without summing an infinite series. Essentially, the FTC states that if a function is continuous over an interval, then the area under the curve (the integral) is related to the antiderivative of the function.

To put it into perspective with the given exercise, once we've expressed the limit of a Riemann sum as a definite integral, the FTC can be applied. For the function \(f(x) = x - x^2\), FTC tells us that to compute the integral from 0 to 1, we can find the antiderivative, which is \(F(x) = \frac{x^2}{2} - \frac{x^3}{3}\), and evaluate it at the end points of the interval. The difference, \(F(1) - F(0)\), gives the integral's value, turning a potentially complex limit process into a straightforward calculation.

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Most popular questions from this chapter

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