91Ó°ÊÓ

The area of a triangle is an essential concept not just in geometry, but also in calculus when dealing with linear functions. Calculating the area of a triangle can help solve integral problems when the function being integrated forms a geometric shape.
To find the area of a triangle, you can use the formula: \ Area = \frac{1}{2} \times \text{base} \times \text{height} \.
Here’s how it works:

• The base is the length of one side of the triangle, usually one that lies along a reference axis.
• The height is the perpendicular distance from the base to the opposite vertex.

In the exercise, the linear function formed a right-angled triangle with the x-axis as the base from \(x = 0\) to \(x = 5\). That makes the base 5 units long.
The height at \(x = 0\) where \(y = 5\) sets the height as 5 units. Plug these into the formula to find the area, yielding an area of 12.5 square units.

The concept of a definite integral is crucial in calculus. It allows us to find the accumulation of quantities, like area, between certain bounds.
When you calculate a definite integral, you are effectively finding the net area between a function and the x-axis over a set interval. For our specific problem, the function is \(5 - x\), and the interval is from 0 to 5.
The definite integral symbol \( \int \) indicates that you are summing up infinitesimal slices of the area under the curve between those limits.
Important points about definite integrals include:

• The result can be interpreted as the total accumulated value, such as distance, area, or volume.
• In the context of the problem, it represents the total area under the line \(5 - x\) from \(x = 0\) to \(x = 5\).
• It yields a single numerical value, which is the net area taking into account areas above and below the x-axis.

This is why the definite integral \( \int_{0}^{5}(5-x) \, dx\) equals the area of the triangle calculated earlier, which is 12.5.

Graphing linear functions is a simple yet powerful tool that makes understanding integration easier. A linear function like \(y = 5-x\) translates to a straight line on a graph. Here’s a breakdown of how this is visualized:
Begin with identifying the basic form of a linear equation \(y = mx + c\), where \(m\) is the slope, and \(c\) is the y-intercept.
For \(y = 5-x\):

• The slope \(m\) is \(-1\), indicating the line slopes downwards.
• The y-intercept \(c\) is 5, meaning the line crosses the y-axis at \(y=5\).

Plotting this, you'd start at the y-intercept (0, 5) and draw a line with a negative slope (-1). This slope tells us that for every 1 unit increase in \(x\), \(y\) decreases by 1 until \(x = 5\) where \(y = 0\).
By graphing, we visually confirmed the structure of the triangle from the origin (0,0) up, along the y-axis to (0,5), then down to (5,0), completing the triangle with the x-axis.