Problem 74 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 73-76, first recognize the given limit as a definite integral and then evaluate that integral by the Second Fundamental Theorem of Calculus \(\lim _{n \rightarrow \infty} \sum_{i=1}^{n}\left(\frac{2 i}{n}\right)^{3} \frac{2}{n}\)

Step-by-Step Solution

Answer

The value of the integral is 4.

Step 1: Identify the Function and Interval

Recognize the given limit as a definite integral by comparing it to the Riemann sum. Notice here that it resembles the form: \[ \lim _{n \rightarrow \infty} \sum_{i=1}^{n} f(x_i) \Delta x \].Identify \( f(x) = x^3 \) and the interval as \([0, 2]\).

Step 2: Convert the Riemann Sum to Integral

The Riemann sum \( \sum_{i=1}^{n}\left(\frac{2 i}{n}\right)^{3} \frac{2}{n} \) becomes the definite integral:\[ \int_{0}^{2} x^3\, dx \].

Step 3: Apply the Second Fundamental Theorem of Calculus

The Second Fundamental Theorem of Calculus allows us to evaluate the integral by finding the antiderivative and then using the limits:The antiderivative of \( x^3 \) is \( \frac{x^4}{4} \).

Step 4: Evaluate the Integral

Evaluate the integral from 0 to 2:\[ \left[ \frac{x^4}{4} \right]_{0}^{2} = \frac{2^4}{4} - \frac{0^4}{4} = \frac{16}{4} - 0 = 4 \].

Key Concepts

Definite Integral

A definite integral represents the accumulation of quantities, and it can be thought of as the net area under a curve in a given interval on a graph.
In the context of the exercise, the limit of a Riemann sum transforms into a definite integral.

• It is expressed with the integral symbol \( \int \), followed by the function and the interval limits.
• The limits of integration indicate the start and end points on the x-axis, like \[ \int_{0}^{2} x^3 \, dx \].

The definite integral gives a precise value that represents the area under the curve from the lower limit to the upper limit of integration.
This is why we are able to accurately calculate solutions to problems involving continuous data over an interval.

Riemann Sum

A Riemann sum is a method for approximating the total area under a curve.
Think of it as a way to estimate an integral using a series of rectangles.

• Each rectangle's height is determined by the value of the function at a specific point within the subinterval.
• The width of each rectangle, called \( \Delta x \), is the same and is defined by splitting the interval into equal parts.
• The Riemann sum adds up the areas of all these rectangles to get a total approximation.

As \( n \) increases, meaning more rectangles are used in the sum, the approximation becomes more accurate.
In terms of calculus, the Riemann sum is the foundation for defining a definite integral as \( n \), the number of intervals, approaches infinity.

Antiderivative

An antiderivative is essentially the reverse process of differentiation.
While derivatives break down functions to find their rates of change, an antiderivative rebuilds the original function given its derivative.

• An antiderivative of a function \( f(x) \) is a function \( F(x) \) such that \( F'(x) = f(x) \).

In the exercise, we need to find the antiderivative to evaluate the integral:

• Given our function \( f(x) = x^3 \), the antiderivative is \( F(x) = \frac{x^4}{4} \).
• This step is crucial as it directly aids in applying the Fundamental Theorem of Calculus to find the exact area under the curve.

Calculus

Calculus is the branch of mathematics focused on the study of change and motion.
Two primary operations in calculus are differentiation (finding the rate of change) and integration (finding the total accumulation).

• Differentiation helps us understand how a function changes at any given point.
• Integration, on the other hand, compiles these changes over an interval.

The Second Fundamental Theorem of Calculus bridges these two concepts.
It states that if you first find an antiderivative of a function and then use the integral limits to evaluate it, you can determine the accumulation of that function over an interval.
This theorem is a powerful tool in calculus, simplifying many complex problems into manageable calculations.