Chapter 5: Problem 76
first recognize the given limit as a definite integral and then evaluate that integral by the Second Fundamental Theorem of Calculus. $$ \lim _{n \rightarrow \infty} \sum_{i=1}^{n}\left[1+\frac{2 i}{n}+\left(\frac{2 i}{n}\right)^{2}\right] \frac{2}{n} $$
Short Answer
Expert verified
The integral evaluates to \( \frac{20}{3} \).
Step by step solution
01
Recognize the Limit as a Riemann Sum
The given limit is of the form \( \lim_{n \to \infty} \sum_{i=1}^{n} f\left(x_i^*\right) \Delta x \). In this case, \( \Delta x = \frac{2}{n} \) and \( x_i^* = \frac{2i}{n} \). The expression inside the sum is \( 1 + x_i^* + (x_i^*)^2 \). Therefore, the function is \( f(x) = 1 + x + x^2 \). The sum is a Riemann sum for the integral of \( f(x) \) over the interval \([0, 2]\).
02
Write the Definite Integral
From the Riemann sum, we identify the definite integral as \( \int_{0}^{2} (1 + x + x^2)\, dx \). This represents the area under the curve \( y = 1 + x + x^2 \) between \( x = 0 \) and \( x = 2 \).
03
Apply the Second Fundamental Theorem of Calculus
To evaluate the integral \( \int_{0}^{2} (1 + x + x^2)\, dx \), we first find an antiderivative of \( f(x) = 1 + x + x^2 \). The antiderivative is \( F(x) = x + \frac{x^2}{2} + \frac{x^3}{3} \).
04
Evaluate the Antiderivative at the Bounds
Using the Second Fundamental Theorem of Calculus, we evaluate the antiderivative at the upper and lower bounds: \[ F(2) - F(0) = \left( 2 + \frac{2^2}{2} + \frac{2^3}{3} \right) - \left( 0 + \frac{0^2}{2} + \frac{0^3}{3} \right) = 2 + 2 + \frac{8}{3}. \]
05
Simplify the Expression
Simplify the expression \( 2 + 2 + \frac{8}{3} \) to get \( 4 + \frac{8}{3} = \frac{12}{3} + \frac{8}{3} = \frac{20}{3} \). Thus, the value of the integral is \( \frac{20}{3} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Riemann Sum
A Riemann Sum is a method for estimating the total area underneath a curve on a graph, otherwise known as the integral of a function over an interval. In our problem, we are given a limit form that is a Riemann sum. It is a sum of the form \( \lim_{n \to \infty} \sum_{i=1}^{n} f\left(x_i^*\right) \Delta x \). Here, \( \Delta x = \frac{2}{n} \) represents the width of each sub-interval, and \( x_i^* = \frac{2i}{n} \) is a point within each sub-interval used to evaluate the function.
- In the exercise, the expression inside the sum is \( 1 + x_i^* + (x_i^*)^2 \).
- This denotes the function \( f(x) = 1 + x + x^2 \) over the interval \([0, 2]\).
Second Fundamental Theorem of Calculus
The Second Fundamental Theorem of Calculus is a cornerstone in the world of integrals. It bridges the concept of differentiation with that of integration. It tells us that if \( F \) is an antiderivative of \( f \) on an interval \( [a, b] \), then:\[\int_{a}^{b} f(x)\,dx = F(b) - F(a)\]
- This theorem allows us to compute definite integrals, transforming the problem of finding areas under curves into a simple subtraction of anti-derivates at boundaries.
- In our exercise, we use it to evaluate the integral \( \int_{0}^{2} (1 + x + x^2) \, dx \).
Antiderivative
Understanding antiderivatives is essential to solving integrals. An antiderivative of a function \( f(x) \) is a function \( F(x) \) such that the derivative of \( F \) is \( f \). In simpler terms, it is the opposite of differentiation.
- In the solution, we find the antiderivative of \( f(x) = 1 + x + x^2 \), which is \( F(x) = x + \frac{x^2}{2} + \frac{x^3}{3} \).
- Each term is integrated separately: \(1\), \(x\), and \(x^2\).
Limit Evaluation
Limit evaluation is a pivotal concept in calculus, particularly in the transition from a Riemann sum to a definite integral. Limits allow us to find values that functions approach as the input gets closer to a particular point or infinity.
- In the case of the given problem, as \( n \to \infty \), the number of sub-intervals increases, reducing \( \Delta x \) to an infinitesimally small size, thus forming a continuous integration.
- This move from the discrete sum (Riemann sum) to the continuous integral requires evaluating the limits precisely.