Problem 53 - Calculus

Question

In Problems \(53-58\), find the area of the region under the curve \(y=f(x)\) over the interval \([a, b] .\) To do this, divide the interval \([a, b]\) into \(n\) equal subintervals, calculate the area of the corresponding circumscribed polygon, and then let \(n \rightarrow \infty\). (See the example for \(y=x^{2}\) in the text.) \(y=x+2 ; a=0, b=1\)

Step-by-Step Solution

Answer

The area is 2.5 square units.

Step 1: Define the Function and Interval

We need to find the area under the curve of \( y = x + 2 \) over the interval \([0, 1]\). This means we are evaluating the area under this line from \( x = 0 \) to \( x = 1 \).

Step 2: Integrate the Function

The area under the curve for a given function over a specific interval is found using definite integration. We need to calculate the integral \( \int_0^1 (x+2) \, dx \).

Step 3: Set up the Integral Expression

Write down the integral: \[ A = \int_0^1 (x + 2) \, dx \]. This represents the accumulated area under the curve from \( x = 0 \) to \( x = 1 \).

Step 4: Evaluate the Integral

Integrate the function: \( \int (x + 2) \, dx = \frac{x^2}{2} + 2x \). Next, substitute the bounds of the integral to find: \[ A = \left[ \frac{x^2}{2} + 2x \right]_0^1 \].

Step 5: Calculate the Definite Integral

Evaluate the expression from Step 4: \[ A = \left( \frac{1^2}{2} + 2 \times 1 \right) - \left( \frac{0^2}{2} + 2 \times 0 \right) \].

Step 6: Simplify the Expression

Calculate the terms: \( \frac{1}{2} + 2 = 2.5 \) and \( 0 \). So, \( A = 2.5 - 0 = 2.5 \).

Step 7: State the Final Result

The area under the curve \( y = x + 2 \) from \( x = 0 \) to \( x = 1 \) is \( 2.5 \).

Key Concepts

Area Under the Curve

When we talk about finding the area under a curve, we're usually referring to a continuous function graphed on a coordinate plane. The region between the curve, the x-axis, and the vertical lines at the endpoints of a specific interval forms the shape whose area we want to calculate. For instance, if you have a curve represented by the function \( y = x + 2 \) and an interval \([0, 1]\), the goal is to find the total area enclosed by the line, the x-axis, and the lines \(x = 0\) and \(x = 1\).

This concept is foundational in calculus and has practical applications in various fields, such as physics, engineering, and economics, where calculating the exact area under a curve is essential for finding quantities such as distances, total change, and accumulated values.

Definite Integral Calculation

The calculation of a definite integral is the process used to find the exact area under a curve over a specified interval. For the function \( y = x + 2 \) over the interval \([0, 1]\), the definite integral represents the sum of the areas of infinitely small rectangles that fit under the curve. The expression is denoted by \( \int_0^1 (x + 2) \, dx \).

• \( dx \) represents an infinitesimally small change in \( x \).
• \( \int \) is the integral symbol, indicating the accumulation of areas over the interval.
• The limits of integration (0 and 1 in this case) define the interval over which we are finding the area.

In practical scenarios, a definite integral not only calculates area but can also represent phenomena like total displacement when position changes over time, or total growth in a population study.

Integral Evaluation Steps

To solve an integral, especially a definite integral, following a systematic approach ensures accuracy and comprehension. For the function \( y = x+ 2 \), these steps are necessary:

- **Define the Function and Interval**: Identify the function \( y = x + 2 \) and the interval \([0, 1]\).
- **Set Up the Integral Expression**: In this scenario, we write down \( \int_0^1 (x + 2) \, dx \).
- **Evaluate the Integral**: Solve the integral by determining its antiderivative. Here, \( \int (x + 2) \, dx = \frac{x^2}{2} + 2x \).
- **Apply the Bounds**: Plug the upper and lower limits into the antiderivative: \( \left[ \frac{x^2}{2} + 2x \right]_0^1 \).
- **Calculate the Result**: Subtract the lower bound evaluation from the upper bound: \( A = \left( \frac{1^2}{2} + 2 \times 1 \right) - \left( \frac{0^2}{2} + 2 \times 0 \right) \), which simplifies to \( 2.5 \).

Following these steps systematically will ensure that you achieve the correct area under the curve, providing a clear understanding of how definite integrals function in practical calculations.