Problem 19 - Algebra 2

Question

Find the area under each curve for the domain \(0 \leq x \leq 1\) $$ f(x)=x+2 $$

Step-by-Step Solution

Answer

The area under the curve \(f(x) = x + 2\) for the domain \(0 \leq x \leq 1\) is 2.5 square units.

Step 1: Understand the problem

The problem asks for the area under the curve defined by the function \(f(x) = x + 2\) between \(x = 0\) and \(x = 1\). By recalling that the area under the curve can be calculated as the definite integral of the function between the given limits, the problem reduces to evaluating \(\int_{0}^{1} (x + 2) dx\).

Step 2: Evaluate the indefinite integral

Before the bounds of the integral are applied, various integration rules can be utilized to compute the indefinite integral of \(x + 2\). The integration is performed term-by-term as follows: \(\int (x + 2) dx = \int x dx + \int2 dx = \frac{1}{2}x^{2} + 2x + C\), where \(C\) is the constant of integration.

Step 3: Apply the Fundametal Theorem of Calculus

To find the definite integral, the Fundamental Theorem of Calculus is applied, which states that the definite integral from a to b of a function is the antiderivative evaluated at b minus the antiderivative evaluated at a. The value of the definite integral from 0 to 1 of the function is found as \(\int_{0}^{1} (x + 2) dx = [\frac{1}{2}x^{2} + 2x]_{0}^{1} = \frac{1}{2}(1)^{2} + 2(1) - [\frac{1}{2}(0)^{2} + 2(0)] = \frac{1}{2} + 2 - 0 = 2.5\).

Step 4: Result and interpretation

The resultant definite integral is \(2.5\), which represents the area under the curve \(y = x + 2\) from \(x = 0\) to \(x = 1\). This solution signifies that in the Cartesian coordinate plane, the area surrounded by the x-axis, the vertical lines \(x = 0\) and \(x = 1\), and the graph of the function \(y = x + 2\) is 2.5 square units.

Key Concepts

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a bridge between differentiation and integration, two core concepts in calculus. This theorem tells us how to compute a definite integral using its related antiderivative function. It essentially makes the complex task of finding the area under a curve merely a subtraction problem.

Here's what the theorem states: If you have a function, say \( f(x) \), that is continuous over an interval \([a, b]\), and \( F(x) \) is an antiderivative of \( f(x) \), then the definite integral of \( f(x) \) from \( a \) to \( b \) is:

\[ \int_{a}^{b} f(x) \, dx = F(b) - F(a) \]

This means you find the antiderivative of \( f(x) \), substitute the limits \( a \) and \( b \), and subtract. It beautifully simplifies what once seemed like an arduous task.

indefinite integral

An indefinite integral, also known as an antiderivative, is a function that reverses differentiation. When you take the indefinite integral of a function, you find another function whose derivative gives you back the original function. When you integrate, you add a constant \( C \) because many different functions can have the same derivative due to the constant term vanishing during differentiation.

Consider the function \( f(x) = x + 2 \). Its indefinite integral is:

• \( \int (x + 2) \ dx = \frac{1}{2} x^2 + 2x + C \)

The process involves integrating term by term.
For \( x \), the integral becomes \( \frac{1}{2}x^2 \), and for the constant \( 2 \), it becomes \( 2x \).

The constant \( C \) is crucial in indefinite integrals, representing a constant that can be any real number.

area under a curve

Finding the area under a curve is a common application of the definite integral. This area provides a visual representation of the integral's value and is important in various fields from physics to economics.

To find this area for the function \( f(x) = x + 2 \) between \( x = 0 \) and \( x = 1 \):

• First, consider the indefinite integral: \( \frac{1}{2} x^2 + 2x + C \).
• Apply the Fundamental Theorem of Calculus with the given limits.
• Calculate \( \int_{0}^{1} (x + 2) \, dx = \frac{1}{2}(1)^2 + 2(1) - [\frac{1}{2}(0)^2 + 2(0)] \).
• This results in the area: \( 2.5 \) square units.

The definite integral helps capture this area perfectly and is calculated by evaluating the antiderivative at the bounds of the interval, then subtracting.