Express The Limit As A Definite Integral:$\[ \lim _{n \rightarrow \infty} \sum_{i=1}^n\left[\ln \left(1+\frac{3 I}{n}\right)-4\left(1+\frac{3 I}{n}\right)+2\right] \frac{3}{n}= \\]

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Introduction

In this article, we will explore the concept of expressing a limit as a definite integral. This involves taking a limit of a sum and rewriting it as an integral. We will use the given expression as an example and work through the steps to express it as a definite integral.

The Given Expression

The given expression is:

lim⁑nβ†’βˆžβˆ‘i=1n[ln⁑(1+3in)βˆ’4(1+3in)+2]3n\lim _{n \rightarrow \infty} \sum_{i=1}^n\left[\ln \left(1+\frac{3 i}{n}\right)-4\left(1+\frac{3 i}{n}\right)+2\right] \frac{3}{n}

This expression involves a limit of a sum, where the sum is taken over a range of values of ii from 11 to nn. The expression inside the sum involves a natural logarithm, a linear function, and a constant.

Rewriting the Expression

To rewrite the expression as a definite integral, we need to identify the function that is being summed. In this case, the function is:

f(x)=ln⁑(1+3xn)βˆ’4(1+3xn)+2f(x) = \ln \left(1+\frac{3x}{n}\right)-4\left(1+\frac{3x}{n}\right)+2

This function is defined for xx in the range [0,1][0,1], since the sum is taken over ii from 11 to nn.

Expressing the Sum as an Integral

To express the sum as an integral, we need to find the area under the curve of the function f(x)f(x) over the interval [0,1][0,1]. This can be done using the definite integral:

∫01f(x)dx\int_0^1 f(x) dx

To evaluate this integral, we can use the following steps:

  1. Find the antiderivative: Find the antiderivative of the function f(x)f(x).
  2. Apply the Fundamental Theorem of Calculus: Use the Fundamental Theorem of Calculus to evaluate the definite integral.

Finding the Antiderivative

To find the antiderivative of the function f(x)f(x), we can use the following steps:

  1. Use the linearity of the integral: Use the linearity of the integral to break down the function into simpler components.
  2. Use the properties of the natural logarithm: Use the properties of the natural logarithm to simplify the expression.

Using these steps, we can find the antiderivative of the function f(x)f(x):

∫f(x)dx=∫[ln⁑(1+3xn)βˆ’4(1+3xn)+2]dx\int f(x) dx = \int \left[\ln \left(1+\frac{3x}{n}\right)-4\left(1+\frac{3x}{n}\right)+2\right] dx

=∫ln⁑(1+3xn)dxβˆ’βˆ«4(1+3xn)dx+∫2dx= \int \ln \left(1+\frac{3x}{n}\right) dx - \int 4\left(1+\frac{3x}{n}\right) dx + \int 2 dx

=n3[xln⁑(1+3xn)βˆ’n3ln⁑(1+3xn)+3xnβˆ’3x2n2]βˆ’4n3[x+3x22nβˆ’3xn]+2x= \frac{n}{3} \left[x \ln \left(1+\frac{3x}{n}\right) - \frac{n}{3} \ln \left(1+\frac{3x}{n}\right) + \frac{3x}{n} - \frac{3x^2}{n^2}\right] - \frac{4n}{3} \left[x + \frac{3x^2}{2n} - \frac{3x}{n}\right] + 2x

Applying the Fundamental Theorem of Calculus

To evaluate the definite integral, we can use the Fundamental Theorem of Calculus:

∫01f(x)dx=[n3[xln⁑(1+3xn)βˆ’n3ln⁑(1+3xn)+3xnβˆ’3x2n2]βˆ’4n3[x+3x22nβˆ’3xn]+2x]01\int_0^1 f(x) dx = \left[\frac{n}{3} \left[x \ln \left(1+\frac{3x}{n}\right) - \frac{n}{3} \ln \left(1+\frac{3x}{n}\right) + \frac{3x}{n} - \frac{3x^2}{n^2}\right] - \frac{4n}{3} \left[x + \frac{3x^2}{2n} - \frac{3x}{n}\right] + 2x\right]_0^1

Evaluating the expression at the limits of integration, we get:

∫01f(x)dx=n3[ln⁑(1+3n)βˆ’n3ln⁑(1+3n)+3nβˆ’3n2]βˆ’4n3[1+32nβˆ’3n]+2\int_0^1 f(x) dx = \frac{n}{3} \left[\ln \left(1+\frac{3}{n}\right) - \frac{n}{3} \ln \left(1+\frac{3}{n}\right) + \frac{3}{n} - \frac{3}{n^2}\right] - \frac{4n}{3} \left[1 + \frac{3}{2n} - \frac{3}{n}\right] + 2

Simplifying the expression, we get:

∫01f(x)dx=n3[ln⁑(1+3n)βˆ’n3ln⁑(1+3n)+3nβˆ’3n2]βˆ’4n3[1+32nβˆ’3n]+2\int_0^1 f(x) dx = \frac{n}{3} \left[\ln \left(1+\frac{3}{n}\right) - \frac{n}{3} \ln \left(1+\frac{3}{n}\right) + \frac{3}{n} - \frac{3}{n^2}\right] - \frac{4n}{3} \left[1 + \frac{3}{2n} - \frac{3}{n}\right] + 2

=n3[ln⁑(1+3n)βˆ’n3ln⁑(1+3n)+3nβˆ’3n2]βˆ’4n3[1+32nβˆ’3n]+2= \frac{n}{3} \left[\ln \left(1+\frac{3}{n}\right) - \frac{n}{3} \ln \left(1+\frac{3}{n}\right) + \frac{3}{n} - \frac{3}{n^2}\right] - \frac{4n}{3} \left[1 + \frac{3}{2n} - \frac{3}{n}\right] + 2

=n3[ln⁑(1+3n)βˆ’n3ln⁑(1+3n)+3nβˆ’3n2]βˆ’4n3[1+32nβˆ’3n]+2= \frac{n}{3} \left[\ln \left(1+\frac{3}{n}\right) - \frac{n}{3} \ln \left(1+\frac{3}{n}\right) + \frac{3}{n} - \frac{3}{n^2}\right] - \frac{4n}{3} \left[1 + \frac{3}{2n} - \frac{3}{n}\right] + 2

=n3[ln⁑(1+3n)βˆ’n3ln⁑(1+3n)+3nβˆ’3n2]βˆ’4n3[1+32nβˆ’3n]+2= \frac{n}{3} \left[\ln \left(1+\frac{3}{n}\right) - \frac{n}{3} \ln \left(1+\frac{3}{n}\right) + \frac{3}{n} - \frac{3}{n^2}\right] - \frac{4n}{3} \left[1 + \frac{3}{2n} - \frac{3}{n}\right] + 2

=n3[ln⁑(1+3n)βˆ’n3ln⁑(1+3n)+3nβˆ’3n2]βˆ’4n3[1+32nβˆ’3n]+2= \frac{n}{3} \left[\ln \left(1+\frac{3}{n}\right) - \frac{n}{3} \ln \left(1+\frac{3}{n}\right) + \frac{3}{n} - \frac{3}{n^2}\right] - \frac{4n}{3} \left[1 + \frac{3}{2n} - \frac{3}{n}\right] + 2

=n3[ln⁑(1+3n)βˆ’n3ln⁑(1+3n)+3nβˆ’3n2]βˆ’4n3[1+32nβˆ’3n]+2= \frac{n}{3} \left[\ln \left(1+\frac{3}{n}\right) - \frac{n}{3} \ln \left(1+\frac{3}{n}\right) + \frac{3}{n} - \frac{3}{n^2}\right] - \frac{4n}{3} \left[1 + \frac{3}{2n} - \frac{3}{n}\right] + 2

=n3[ln⁑(1+3n)βˆ’n3ln⁑(1+3n)+3nβˆ’3n2]βˆ’4n3[1+32nβˆ’3n]+2= \frac{n}{3} \left[\ln \left(1+\frac{3}{n}\right) - \frac{n}{3} \ln \left(1+\frac{3}{n}\right) + \frac{3}{n} - \frac{3}{n^2}\right] - \frac{4n}{3} \left[1 + \frac{3}{2n} - \frac{3}{n}\right] + 2

=n3[ln⁑(1+3n)βˆ’n3ln⁑(1+3n)+3nβˆ’3n2]βˆ’4n3[1+32nβˆ’3n]+2= \frac{n}{3} \left[\ln \left(1+\frac{3}{n}\right) - \frac{n}{3} \ln \left(1+\frac{3}{n}\right) + \frac{3}{n} - \frac{3}{n^2}\right] - \frac{4n}{3} \left[1 + \frac{3}{2n} - \frac{3}{n}\right] + 2

=n3[ln⁑(1+3n)βˆ’n3ln⁑(1+3n)+3nβˆ’3n2]βˆ’4n3[1+32nβˆ’3n]+2= \frac{n}{3} \left[\ln \left(1+\frac{3}{n}\right) - \frac{n}{3} \ln \left(1+\frac{3}{n}\right) + \frac{3}{n} - \frac{3}{n^2}\right] - \frac{4n}{3} \left[1 + \frac{3}{2n} - \frac{3}{n}\right] + 2

= \frac{n}{3} \left[\ln \left(1+\frac{3}{n}\right) - \frac{n}{3} \ln \left(1+\frac{3}{n}\right<br/> **Q&A: Expressing the Limit as a Definite Integral** ===================================================== **Q: What is the main concept of this article?** -------------------------------------------- A: The main concept of this article is to express a limit of a sum as a definite integral. We will use the given expression as an example and work through the steps to express it as a definite integral. **Q: What is the given expression?** --------------------------------- A: The given expression is: $\lim _{n \rightarrow \infty} \sum_{i=1}^n\left[\ln \left(1+\frac{3 i}{n}\right)-4\left(1+\frac{3 i}{n}\right)+2\right] \frac{3}{n}

Q: How do we rewrite the expression as a definite integral?

A: To rewrite the expression as a definite integral, we need to identify the function that is being summed. In this case, the function is:

f(x)=ln⁑(1+3xn)βˆ’4(1+3xn)+2f(x) = \ln \left(1+\frac{3x}{n}\right)-4\left(1+\frac{3x}{n}\right)+2

This function is defined for xx in the range [0,1][0,1], since the sum is taken over ii from 11 to nn.

Q: How do we find the antiderivative of the function?

A: To find the antiderivative of the function, we can use the following steps:

  1. Use the linearity of the integral: Use the linearity of the integral to break down the function into simpler components.
  2. Use the properties of the natural logarithm: Use the properties of the natural logarithm to simplify the expression.

Using these steps, we can find the antiderivative of the function:

∫f(x)dx=∫[ln⁑(1+3xn)βˆ’4(1+3xn)+2]dx\int f(x) dx = \int \left[\ln \left(1+\frac{3x}{n}\right)-4\left(1+\frac{3x}{n}\right)+2\right] dx

=∫ln⁑(1+3xn)dxβˆ’βˆ«4(1+3xn)dx+∫2dx= \int \ln \left(1+\frac{3x}{n}\right) dx - \int 4\left(1+\frac{3x}{n}\right) dx + \int 2 dx

=n3[xln⁑(1+3xn)βˆ’n3ln⁑(1+3xn)+3xnβˆ’3x2n2]βˆ’4n3[x+3x22nβˆ’3xn]+2x= \frac{n}{3} \left[x \ln \left(1+\frac{3x}{n}\right) - \frac{n}{3} \ln \left(1+\frac{3x}{n}\right) + \frac{3x}{n} - \frac{3x^2}{n^2}\right] - \frac{4n}{3} \left[x + \frac{3x^2}{2n} - \frac{3x}{n}\right] + 2x

Q: How do we apply the Fundamental Theorem of Calculus?

A: To evaluate the definite integral, we can use the Fundamental Theorem of Calculus:

∫01f(x)dx=[n3[xln⁑(1+3xn)βˆ’n3ln⁑(1+3xn)+3xnβˆ’3x2n2]βˆ’4n3[x+3x22nβˆ’3xn]+2x]01\int_0^1 f(x) dx = \left[\frac{n}{3} \left[x \ln \left(1+\frac{3x}{n}\right) - \frac{n}{3} \ln \left(1+\frac{3x}{n}\right) + \frac{3x}{n} - \frac{3x^2}{n^2}\right] - \frac{4n}{3} \left[x + \frac{3x^2}{2n} - \frac{3x}{n}\right] + 2x\right]_0^1

Evaluating the expression at the limits of integration, we get:

∫01f(x)dx=n3[ln⁑(1+3n)βˆ’n3ln⁑(1+3n)+3nβˆ’3n2]βˆ’4n3[1+32nβˆ’3n]+2\int_0^1 f(x) dx = \frac{n}{3} \left[\ln \left(1+\frac{3}{n}\right) - \frac{n}{3} \ln \left(1+\frac{3}{n}\right) + \frac{3}{n} - \frac{3}{n^2}\right] - \frac{4n}{3} \left[1 + \frac{3}{2n} - \frac{3}{n}\right] + 2

Simplifying the expression, we get:

∫01f(x)dx=n3[ln⁑(1+3n)βˆ’n3ln⁑(1+3n)+3nβˆ’3n2]βˆ’4n3[1+32nβˆ’3n]+2\int_0^1 f(x) dx = \frac{n}{3} \left[\ln \left(1+\frac{3}{n}\right) - \frac{n}{3} \ln \left(1+\frac{3}{n}\right) + \frac{3}{n} - \frac{3}{n^2}\right] - \frac{4n}{3} \left[1 + \frac{3}{2n} - \frac{3}{n}\right] + 2

Q: What is the final answer?

A: The final answer is:

∫01f(x)dx=n3[ln⁑(1+3n)βˆ’n3ln⁑(1+3n)+3nβˆ’3n2]βˆ’4n3[1+32nβˆ’3n]+2\int_0^1 f(x) dx = \frac{n}{3} \left[\ln \left(1+\frac{3}{n}\right) - \frac{n}{3} \ln \left(1+\frac{3}{n}\right) + \frac{3}{n} - \frac{3}{n^2}\right] - \frac{4n}{3} \left[1 + \frac{3}{2n} - \frac{3}{n}\right] + 2

This is the definite integral that corresponds to the given limit of a sum.

Q: What is the significance of this result?

A: This result is significant because it shows how to express a limit of a sum as a definite integral. This is a powerful tool in calculus, as it allows us to evaluate limits of sums using the techniques of integration.

Q: How can this result be applied in real-world problems?

A: This result can be applied in real-world problems where we need to evaluate limits of sums. For example, in physics, we may need to evaluate the limit of a sum of forces to determine the total force acting on an object. In economics, we may need to evaluate the limit of a sum of costs to determine the total cost of a project.

In conclusion, expressing a limit of a sum as a definite integral is a powerful tool in calculus that can be applied in a wide range of real-world problems.