Problem 43 - Essential Calculus

Question

In Exercises \(43-48,\) use the limit process to find the area of the region between the graph of the function and the \(x\) -axis over the given interval. Sketch the region. $$ y=-2 x+3, \quad[0,1] $$

Step-by-Step Solution

Answer

The area of the region between the graph of the function \(y = -2x + 3\) and the \(x\) -axis over the interval [0,1] is 2 square units.

Step 1: Sketch the Region

Start by graphing this linear function on the interval [0,1]. This line intersects the x-axis around 1.5, so on [0,1] it lies completely above the x-axis.

Step 2: Setup the Integral

Since we are finding the area between the function and the x-axis over the interval [0,1], setup the integral as follows: \(\int_{0}^{1} (-2x + 3)dx\).

Step 3: Evaluate the Integral

To evaluate this definite integral, first find the antiderivative of the integrand. An antiderivative of \(-2x + 3\) is \(-x^2 + 3x\). Then, apply the Second Fundamental Theorem of Calculus by subtracting the antiderivative evaluated at 0 from the antiderivative evaluated at 1. This gives \((-(1)^2 + 3 * (1)) - (-(0)^2 + 3 * (0)) = 2\).

Key Concepts

Limit Process

The limit process is a foundational concept in calculus, helping us to understand how we approach the area under a curve. In problems like the one involving the function \(y = -2x + 3\), we can use the limit process to find the area between the curve and the x-axis over a specified interval, in this case, \([0,1]\).
To use the limit process, imagine slicing the area into countless thin rectangles. The height of each rectangle is determined by the function's value at a given \(x\), and the width of each is a tiny increment, \(\Delta x\). As the number of rectangles increases, and \(\Delta x\) approaches zero, we move towards the true area.

• Set up a sum of the areas of these rectangles.
• The sum transitions into an integral as we let the number of rectangles approach infinity.

In this exercise, the sum becomes the integral \(\int_{0}^{1} (-2x + 3) \, dx\), indicating we're ready to find the area.

Definite Integral

A definite integral represents the signed area under a curve within a given interval on the x-axis. For the function \(y = -2x + 3\), the definite integral from 0 to 1 calculates the exact area between the curve and the x-axis along this interval.
Definite integrals are crucial because:

• They allow us to calculate the net area, considering areas above the x-axis as positive and those below as negative.
• They provide a concise way to express the total accumulation of quantities, such as distance, area, or mass.

For our problem, the integral is \(\int_{0}^{1} (-2x + 3) \, dx\). Evaluating this integral gives us the precise area, which is a helpful method to translate problems about physical quantities into mathematical expressions.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus bridges the gap between derivatives and integrals, offering a strategy to evaluate definite integrals such as \(\int_{0}^{1} (-2x + 3) \, dx\). It connects antiderivatives with the accumulated area under a curve.
This theorem has two main parts:

• The first part states that if a function is continuous over an interval, then its antiderivative can be used to determine the integral over that interval.
• The second part, more relevant here, specifically indicates that the integral from \(a\) to \(b\) of a function is simply the difference between the values of its antiderivative at \(b\) and \(a\).

In our exercise, we find the antiderivative of \(-2x + 3\), which is \(-x^2 + 3x\). We then evaluate this antiderivative at the boundaries of our interval: \(1\) and \(0\). Plug these in and subtract, resulting in the value \(2\), confirming the area under the curve over \([0,1]\). This theorem simplifies what would otherwise be an arduous computation process.