Problem 25 - Algebra 2

Question

Find the area under each curve for the domain \(0 \leq x \leq 1\) $$ y=4 x+2 $$

Step-by-Step Solution

Answer

The area under the curve from \(x = 0\) to \(x = 1\) is 4 square units.

Step 1: Understanding the Problem

We are required to find the area under the curve of the function \(y = 4x + 2\) from \(x = 0\) to \(x = 1\). This can be accomplished by computing the definite integral of the function over the given interval.

Step 2: Setting up the Integral

The area under the curve can be found by calculating the definite integral of the function within the limits of the domain. So we set up the integral as follows: \[\int_0^1 (4x + 2)dx\]

Step 3: Calculating the Integral

The integral of \(4x\) is \(2x^2\) and the integral of \(2\) is \(2x\). Applying the power rule and the constant multiple rule of integration, we find the integral of the function: \[\int_0^1 (4x + 2)dx = \int_0^1 4x dx + \int_0^1 2 dx = 2x^2 + 2x \bigg|_0^1\]

Step 4: Evaluating the Integral

We plug in the upper and lower limits of the integral: \[2(1)^2 + 2(1) - (2(0)^2 + 2(0)) = 2 + 2 - 0 = 4\].

Key Concepts

Integration in Algebra

Integration is a fundamental concept in algebra that serves as a powerful tool for solving a variety of mathematical problems, such as finding areas, volumes, and average values. In the context of algebra, integration can be thought of as the process of calculating the area under a curve on a graph, particularly when the curve represents a function.

When we integrate a function, we essentially add up an infinite number of infinitesimally small rectangles under the curve to find the total area. Integration is the inverse operation of differentiation, meaning that while differentiation gives us the slope of the curve at any given point, integration gives us the accumulation of the curve's values over an interval.

For functions that can be expressed algebraically, such as polynomials, exponentials, or trigonometric functions, there are specific rules and methods to integrate them. In the case of the exercise given, we are required to find the definite integral of a linear function over a specified interval, which is a basic application of integration in algebra.

Area Under a Linear Function

The area under a linear function on a graph can be visualized as the space trapped between the function's line, the x-axis, and the vertical lines drawn at the boundaries of the interval of interest. For the linear function expressed by the equation \(y = 4x + 2\), the graph forms a trapezoid between \(x = 0\) and \(x = 1\).

Calculating the area under a linear function like this is done using integration. The definite integral from \(x = 0\) to \(x = 1\) of the function provides us with the exact area of this trapezoidal shape. By setting up the integral and evaluating it as shown in the solution steps, we have effectively calculated the area under the line on the graph for the given domain.

Power Rule for Integration

The power rule for integration is a basic rule in calculus that simplifies the process of integrating polynomial functions. It dictates that when integrating a term like \(ax^n\), where \(a\) is a coefficient and \(n\) is a real number, the antiderivative is given by \(\frac{a}{n+1}x^{n+1}\), provided that \(n eq -1\).

In our exercise, the function includes a constant term, which is essentially \(x^0\), and a linear term, \(4x\) which is \(x^1\). Applying the power rule for the linear term, we find that the antiderivative is \(2x^2\). For the constant term, since any constant \(c\) can be integrated to \(cx\), the antiderivative is \(2x\). Combining these, we acquire the integrated function over our interval, leading to the solution which represents the definite integral, or the area under the curve.