Problem 74 - Calculus Early Transcendentals: Pearson New International Edition

Question

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 limit equals 4.

Step 1: Recognize the Limit as a Riemann Sum

The given limit \(\lim _{n \rightarrow \infty} \sum_{i=1}^{n}\left(\frac{2 i}{n}\right)^{3} \frac{2}{n}\) resembles the Riemann sum for a definite integral. The form \(\sum_{i=1}^{n} f(x_i^*) \Delta x\) where \(x_i^* = \frac{2i}{n}\) and \(\Delta x = \frac{2}{n}\) corresponds to the function \(f(x) = x^3\) being integrated over the interval \([0, 2]\).

Step 2: Set Up the Definite Integral

The expression \(\lim _{n \rightarrow \infty} \sum_{i=1}^{n} f(x_i^*) \Delta x\) translates to the definite integral notation \[ \int_{0}^{2} x^3 \, dx \] based on the identified function \(f(x) = x^3\) and the limits reflecting the interval \([0, 2]\).

Step 3: Apply the Second Fundamental Theorem of Calculus

The Second Fundamental Theorem of Calculus states that if \(F\) is an antiderivative of \(f\) on an interval \([a, b]\), then \(\int_{a}^{b} f(x) \, dx = F(b) - F(a)\). For \(f(x) = x^3\), an antiderivative is \(F(x) = \frac{x^4}{4}\).

Step 4: Evaluate the Antiderivative at the Bounds

Calculate \(F(2)\) and \(F(0)\) using \(F(x) = \frac{x^4}{4}\): - \(F(2) = \frac{2^4}{4} = \frac{16}{4} = 4\)- \(F(0) = \frac{0^4}{4} = 0\)

Step 5: Calculate the Definite Integral

Compute the definite integral \(\int_{0}^{2} x^3 \, dx\) by subtracting: \[ F(2) - F(0) = 4 - 0 = 4 \] The value of the integral is 4, which is the value of the original limit.

Key Concepts

Riemann sums

Riemann sums are a way to approximate the area under a curve, which sets the foundation for understanding definite integrals. They involve dividing the interval over which a function is defined into small segments, often called subintervals.
In these subintervals, a point is chosen, usually a left endpoint, right endpoint, or the midpoint, to evaluate the function.

• The function value at this point is multiplied by the width of the subinterval, denoted as \( \Delta x \).
• The sum of all these small rectangles' areas approximates the area under the curve.

For the example given in the problem, the Riemann sum \( \lim _{n \rightarrow \infty} \sum_{i=1}^{n}\left(\frac{2 i}{n}\right)^{3} \frac{2}{n} \) approximates the integral of \( x^3 \) over the interval \([0, 2]\).
The expression reveals a typical Riemann sum where each term \( \left(\frac{2 i}{n}\right)^{3} \frac{2}{n} \) represents the area of one of the rectangles under the curve \( f(x) = x^3 \).

Second Fundamental Theorem of Calculus

The Second Fundamental Theorem of Calculus plays a crucial role in bridging the gap between antiderivatives and definite integrals. It asserts that if a function \( f \) is continuous on a closed interval \([a, b]\) and \( F \) is any antiderivative of \( f \) on this interval, then:
\[ \int_{a}^{b} f(x) \, dx = F(b) - F(a) \]

• This theorem simplifies the computation of the definite integral to finding an antiderivative \( F \) and evaluating it at the boundaries \( a \) and \( b \).

In our example, the function \( f(x) = x^3 \) is continuous over the interval \([0, 2]\).
By the theorem, we use an antiderivative of \( f(x) = x^3 \), which is \( F(x) = \frac{x^4}{4} \), and calculate the area under the curve within the given bounds by evaluating \( F(x) \) at \( b = 2 \) and \( a = 0 \).

antiderivatives

Antiderivatives, also known as indefinite integrals, are functions that "undo" the process of differentiation. For any given function \( f(x) \), an antiderivative \( F(x) \) is a function such that \( F'(x) = f(x) \).
This concept is pivotal in solving definite integrals using the Second Fundamental Theorem of Calculus.

• In our example, an antiderivative of \( f(x) = x^3 \) is \( F(x) = \frac{x^4}{4} \).

To find an antiderivative, you can reverse-engineer the differentiation rules:

• For polynomials, increase the power by one and divide by the new power. Thus, for \( x^3 \) the antiderivative is \( \frac{x^4}{4} \).

When applied to definite integrals, antiderivatives provide a straightforward way to determine the area under a curve between specified bounds, simplifying the process considerably.