Problem 1 - Calculus Early Transcendentals: Pearson New International Edition

Question

use the Second Fundamental Theorem of Calculus to evaluate each definite integral. $$ \int_{0}^{2} x^{3} d x $$

Step-by-Step Solution

Answer

The integral evaluates to 4.

Step 1: Identify the Function

The integral we need to evaluate is \( \int_{0}^{2} x^{3} \, dx \). Here, the function \( f(x) = x^3 \) is the integrand.

Step 2: Find the Antiderivative

The antiderivative of \( x^3 \) is \( F(x) = \frac{x^4}{4} + C \), where \( C \) is the constant of integration. For definite integrals, we don't need the constant.

Step 3: Apply the Second Fundamental Theorem of Calculus

According to the Second Fundamental Theorem of Calculus, \( \int_{a}^{b} f(x) \, dx = F(b) - F(a) \). In this case, \( a = 0 \) and \( b = 2 \), so we need to calculate \( F(2) - F(0) \).

Step 4: Evaluate the Antiderivative at the Bounds

Calculate \( F(2) = \frac{2^4}{4} = \frac{16}{4} = 4 \) and \( F(0) = \frac{0^4}{4} = 0 \).

Step 5: Compute the Definite Integral

Subtract the two results: \( F(2) - F(0) = 4 - 0 = 4 \). Thus, the value of the integral \( \int_{0}^{2} x^3 \, dx = 4 \).

Key Concepts

Definite Integral

A definite integral is a concept that represents the area under a curve of a function, between two specified limits or boundaries. In the exercise, this is represented by \( \int_{0}^{2} x^{3} \, dx \). The limits in this context are from \( x=0 \) to \( x=2 \).

• The integral sign \( \int \) signifies the operation of integration.
• The number \( 0 \) is the lower limit of the integral, indicating where we begin considering the area.
• The number \( 2 \) is the upper limit, indicating where we stop considering the area.

We use the Second Fundamental Theorem of Calculus to evaluate definite integrals. It states that if we have an antiderivative, we can compute the area under the curve by evaluating this antiderivative at the upper and lower limits and subtracting.

Antiderivative

An antiderivative is essentially the opposite of taking a derivative. If a function \( f(x) \) has an antiderivative \( F(x) \), it means that the derivative of \( F(x) \) gives us back the function \( f(x) \).
In the context of the exercise, we started with the function \( f(x) = x^3 \). The antiderivative of this function is \( F(x) = \frac{x^4}{4} + C \), where \( C \) is a constant.

• The process of finding an antiderivative is known as integration.
• When performing definite integrals, the constant \( C \) does not affect the result and is thus omitted.

The antiderivative allows us to apply the Second Fundamental Theorem of Calculus, making it essential for evaluating definite integrals.

Integrand

The integrand is simply the function that you are integrating, the function under the integral sign. In the given exercise, the integrand is \( x^3 \). This is the function whose area under its curve we are interested in calculating from \( x=0 \) to \( x=2 \).

• It's important to correctly identify the integrand before proceeding with finding its antiderivative.
• Understanding the behavior of the integrand can give insights into the nature of the integral's result.

In our example, the integrand \( x^3 \) is a simple polynomial function, making it straightforward to find its antiderivative.

Evaluation of Integral

The evaluation of an integral refers to the process of calculating its value over a specific interval, showing the area under the given curve between the specified limits. In the context of definite integrals, we employ the Second Fundamental Theorem of Calculus.
The process involves:

• Calculating the antiderivative \( F(x) \) of the integrand \( f(x) \).
• Substituting the upper limit into \( F(x) \) which gives \( F(b) \), and the lower limit which gives \( F(a) \).
• Finding the difference: \( F(b) - F(a) \).

For the exercise given, after finding the antiderivative, we evaluated it at \( x=2 \) and \( x=0 \), giving \( F(2) = 4 \) and \( F(0) = 0 \). Subtraction leads to \( F(2) - F(0) = 4 - 0 = 4 \), which is the value of the definite integral. This shows the area under the curve \( x^3 \) from 0 to 2.