Problem 13 - A Graphical Approach to Precalculus with Limits

Question

Consider the region below \(f(x)=\frac{x}{2},\) above the \(x\) -axis, and between \(x=0\) and \(x=4 .\) Let \(x_{i}\) be the midpoint of the \(i\) th subinterval. (a) Approximate the area of the region, using four rectangles. (b) Find \(\int_{0}^{4} f(x) d x\) by using the formula for the area of a triangle.

Step-by-Step Solution

Answer

The approximate area using four rectangles is 4, and the exact area using the triangle formula is also 4.

Step 1: Understand the problem

We are asked to approximate the area under the curve \(f(x) = \frac{x}{2}\), above the \(x\)-axis, and between \(x=0\) and \(x=4\) using four rectangles. Then, we need to calculate the exact area by recognizing the shape of the region as a triangle.

Step 2: Divide the interval for approximation

We need to divide the interval \([0, 4]\) into four equal subintervals. Each subinterval has a length \(\Delta x = \frac{4-0}{4} = 1\). So, the points are \(x = 0, 1, 2, 3, 4\).

Step 3: Find the midpoint of each subinterval

For each subinterval \([x_{i-1}, x_i]\), the midpoint is \(x_i = \frac{x_{i-1} + x_i}{2}\). The midpoints are thus at \(x_1 = 0.5\), \(x_2 = 1.5\), \(x_3 = 2.5\), \(x_4 = 3.5\).

Step 4: Evaluate the function at the midpoints

Calculate \(f(x_i)\) for each midpoint: - \(f(0.5) = \frac{0.5}{2} = 0.25\)- \(f(1.5) = \frac{1.5}{2} = 0.75\)- \(f(2.5) = \frac{2.5}{2} = 1.25\)- \(f(3.5) = \frac{3.5}{2} = 1.75\)

Step 5: Calculate the approximate area using rectangles

The area of each rectangle is given by \(f(x_i) \times \Delta x\). The total approximate area is:\[\text{Approximation} = (0.25 \times 1) + (0.75 \times 1) + (1.25 \times 1) + (1.75 \times 1) = 4\]

Step 6: Recognize the shape and calculate the exact area

The region under \(f(x) = \frac{x}{2}\), from \(x = 0\) to \(x = 4\), forms a right triangle with base \(4\) and height \(2\) (at \(x = 4\), \(y = \frac{4}{2} = 2\)). The area of the triangle is:\[\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 2 = 4\]

Step 7: Compare and conclude

Both the approximation and the exact calculation yield an area of 4, confirming the correctness of our midpoint approximation and the triangular area formula.

Key Concepts

Area Approximation

Area approximation comes into play when we want an estimate of the area under a curve, usually when finding an exact value could be complex or time-consuming. In this exercise, we looked at the function \(f(x) = \frac{x}{2}\) over the interval \([0, 4]\). Here, the challenge is to approximate the area above the \(x\)-axis and below the curve using a simple geometric method.
The task involves dividing the region into smaller sections—in this case, rectangles. Each rectangle helps us estimate part of the total area under the curve. By summing the areas of these rectangles, we get a good approximation of the area we are interested in. It's like breaking the problem into smaller, more manageable pieces that can be put back together to form the whole.

• This concept often uses simpler computations instead of complex integration.
• The result isn't exact, but it usually gets close.
• It's a practical way, especially when using numerical methods on computers.

Midpoint Rule

The midpoint rule is one approach for approximating the area under a curve by using rectangles. For each subinterval, rather than choosing the endpoint or any arbitrary point to decide the rectangle's height, we use the midpoint. This technique can sometimes provide better approximations than using the left or right endpoints alone.
In our example, we divided the interval \([0, 4]\) into 4 subintervals, each of length \(\Delta x = 1\). The midpoints were found as \(x_1 = 0.5\), \(x_2 = 1.5\), \(x_3 = 2.5\), and \(x_4 = 3.5\). Evaluating the function at these midpoints allows us to construct rectangles whose heights represent function values at these middle points.

• It provides a balanced method for deploying rectangles.
• Results can closely align with the exact area, especially for smooth functions.
• Considered easy to implement and compute, especially for beginners.

Integral of a Function

The integral of a function offers a way to compute the exact area under a curve, often bypassing complications that arise through different approximation methods. The integral combines these values smoothly and continuously, offering highly accurate results. In this context, we aim to integrate the function \(f(x) = \frac{x}{2}\) from \(x = 0\) to \(x = 4\).
However, for easy calculations without integral calculus, recognizing the region as a geometric shape can prove helpful, as you see in the next section. Integrals essentially sum an infinite number of infinitesimally small sections to give you the total area.

• Integral calculus finds exact areas under curves.
• Applications extend beyond simple geometric shapes.
• Highly useful across various fields, especially sciences and engineering.

Triangle Area Formula

Shapes sometimes make calculating the area simpler. Geometrically, if you can recognize the enclosed area under a graph as a triangle, as in this exercise, it's straightforward to use the formula for the area of a triangle to find the integral.
For the function \(f(x) = \frac{x}{2}\) over \([0, 4]\), the region forms a right triangle with a base along the \(x\)-axis from \(0\) to \(4\), and a height given by \(f(4) = 2\). The area of a right triangle can be calculated using:\[\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}\]Inserting the values, the area becomes \(\frac{1}{2} \times 4 \times 2 = 4\). This matches our approximation perfectly.

• Using geometrical insights can simplify calculations.
• Useful when the interval creates a special shape like a rectangle or triangle.
• Helps in verifying the numerical approximation methods used previously.