Chapter 4: Problem 75
Evaluate the limit by interpreting it as the limit of a Riemann sum of a function on the interval \([a, b]\). $$ \lim _{n \rightarrow \infty} \frac{2}{n} \sum_{k=1}^{n}\left(2+\frac{2 k}{n}\right)^{2} ; \quad[2,4] $$
Short Answer
Expert verified
The limit of the given Riemann sum is the definite integral of the function \(f(x) = (2 + x)^2\) over the interval [2,4]. Evaluating this definite integral, we find that the limit of the Riemann sum as n approaches infinity is 51\(\frac{1}{3}\).
Step by step solution
01
Identify the function inside the Riemann sum
The function is given by:
$$
f(x) = \left(2 + \frac{2k}{n}\right)^2
$$
Replace the discrete variable k with the continuous variable x:
$$
f(x) = (2 + x)^2
$$
02
Set up the definite integral
Now that the function has been found, we can set up the definite integral over the interval [2,4]:
$$
\int_2^4 (2 + x)^2 dx
$$
03
Evaluate the definite integral
To find the value of the definite integral, we need to find the antiderivative of the function and evaluate it at the bounds of the interval:
$$
\int (2 + x)^2 dx = \frac{1}{3}(2 + x)^3 + C
$$
Now, evaluate the antiderivative at the bounds 2 and 4:
$$
\left[\frac{1}{3}(2 + x)^3\right]_2^4 = \frac{1}{3}(2 + 4)^3 - \frac{1}{3}(2 + 2)^3
$$
04
Compute the final result
Calculate the difference using the evaluated bounds to get the definite integral's value:
$$
\frac{1}{3}(2 + 4)^3 - \frac{1}{3}(2 + 2)^3 = \frac{1}{3}(6^3) - \frac{1}{3}(4^3) = \frac{1}{3}(216 - 64)
$$
Now, simplify the expression to get the limit of the Riemann sum:
$$
\lim _{n \rightarrow \infty} \frac{2}{n} \sum_{k=1}^{n}\left(2 + \frac{2 k}{n}\right)^{2}=\frac{1}{3}(216-64)=\frac{1}{3}(152)=51\frac{1}{3}
$$
The limit of the Riemann sum as n approaches infinity is 51\(\frac{1}{3}\).
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Definite Integral
The definite integral is a powerful concept used to find the area under a curve between two defined limits on the x-axis, usually denoted as \([a, b]\). It is an essential tool in calculus for accumulative measurements, such as volume and total distance. A definite integral is represented as \( \int_a^b f(x) \, dx \), where \(f(x)\) is the function being integrated.
- "Definite" signifies that the boundaries \(a\) and \(b\) are specific numbers.
- The variable \(x\) indicates that the integration is performed with respect to \(x\).
Limit of a Function
In calculus, the limit of a function is a fundamental aspect reflecting what value a function approaches as the input approaches a particular point. It's crucial for understanding how functions behave as variables get indefinitely close to a particular value or infinity. Limits underpin much of how calculus is structured, providing a way to deal with continuous change.
- The notation \(\lim_{x \to a} f(x) = L\) expresses that as \(x\) gets arbitrarily close to \(a\), \(f(x)\) approaches \(L\).
- When considering infinite sequences or sums, limits help define and evaluate shapes and areas that are otherwise complex.
Antiderivative
An antiderivative of a function \(f(x)\), often called an "indefinite integral," is another function whose derivative yields \(f(x)\). If \(F'(x) = f(x)\), then \(F(x)\) is the antiderivative of \(f(x)\). Antiderivatives are essential in calculating definite integrals, as they enable us to find the accumulated change between two points.
- Finding an antiderivative is a reversal of differentiation.
- The notation is \( \int f(x)\, dx = F(x) + C \), where \(C\) is the integration constant.
Integration
Integration is the process of finding the integral of a function, essentially extending the concept of areas and volumes to more complex functions and shapes. As opposed to differentiation that focuses on rates of change, integration accumulates total quantities.
- In simple terms, it acts like a sum of tiny parts that make a whole.
- Integration helps compute values such as total distance, area, volume, and even sums of infinite series.