Problem 28 - Calculus Early Transcendentals

Question

Net area and definite integrals Use geometry (not Riemann sums) to evaluate the following definite integrals. Sketch a graph of the integrand, show the region in question, and interpret your result. $$\int_{0}^{2}(1-x) d x$$

Step-by-Step Solution

Answer

Answer: The net area under the curve of the function $f(x) = 1-x$ from $x=0$ to $x=2$ is 1.

Step 1: Sketch the Graph of the Integrant

To start, we'll need to sketch the graph of the function \(f(x) = 1-x\). To do this, first note that it is a linear function with a slope of -1 and a y-intercept of 1. Thus, it will be a line that starts at point \((0,1)\) and moves down and to the right. Sketch this graph on x-axis ranging from 0 to 2.

Step 2: Identify the Region Under the Curve

Next, we need to identify the region that lies between the x-axis and the curve from \(x=0\) to \(x=2\). This region should be a triangle with vertices at \((0,0)\), \((0,1)\), and \((2,0)\).

Step 3: Calculate the Area of the Region

Now that we have identified the region, we can use geometry to calculate its area. The region is a triangle with a base of 2 and a height of 1. The formula for the area of a triangle is given by \(\frac{1}{2} \times base \times height\). Plugging in the values for our region, we get: $$Area = \frac{1}{2} \times 2 \times 1 = 1$$

Step 4: Interpret the Result

The definite integral represents the net area under the curve of the function \((1-x)\) from \(x=0\) to \(x=2\). Using geometry, we found that this net area is 1. This means that the integral \(\int_{0}^{2}(1-x) d x\) evaluates to 1.

Key Concepts

Net Area

Understanding the net area in the context of definite integrals is essential for students tackling calculus problems. The term 'net area' refers to the total signed area between a curve and the x-axis, within a specific range. When calculating this net area using a definite integral, such as with \(\int_{0}^{2}(1-x) dx\), areas above the x-axis are considered positive, while areas below the x-axis are negative.

Here's a simple example to illustrate this concept. For the given problem, we begin with sketching the linear function \(f(x) = 1-x\) to visualize the area we're looking to find. The next step is identifying the area under the curve - in this case, it's a triangle between \(x=0\) and \(x=2\). Lastly, we calculate this triangle's area using the formula \(Area = \frac{1}{2} \times base \times height\), yielding an area of 1. It's crucial to note that because the entire region lies above the x-axis, the net area is also 1, confirming that the integral evaluates to a positive 1, rather than a negative value. Students should become comfortable in distinguishing when to add or subtract areas under the curve to correctly compute the 'net area' for more complex functions.

Geometric Interpretation of Integrals

The geometric interpretation of integrals provides a tangible way to visualize the abstract concept of integration. In calculus, a definite integral represents the accumulation of quantities, which can often be described in terms of areas under curves on graphs. When we use a definite integral, such as \(\int_{a}^{b} f(x) dx\), we're effectively summing up infinitely many infinitesimally small areas under the curve between \(x=a\) and \(x=b\).

For the function \(f(x) = 1-x\), this interpretation translates into finding the area of a geometric shape bound by the graph of the function, the x-axis, and the vertical lines \(x=0\) and \(x=2\). By recognizing this region as a triangle, we apply simple geometry instead of complicated integral calculus methods. This concretizes the abstract concept and emphasizes why integrals are often introduced with the area-under-a-curve analogy. This visual approach aids in comprehending more complex scenarios where we may encounter curves that aren't so neatly represented by basic shapes.

Area Under a Curve

The concept of 'area under a curve' is among the foundational ideas in integral calculus. It is used to determine the size of the region that lies below the graph of a function and above the x-axis. In mathematical terms, if you have a function \(f(x)\) defined on an interval \[a, b\], the area under the curve from \(a\) to \(b\) is given by the definite integral \(\int_{a}^{b} f(x) dx\).

In the exercise with \(f(x) = 1-x\), the graph crosses the x-axis once within our interval of interest. Since the resulting figure is a right triangle, we use the formula for the area of a triangle (half of base times height) to find that the area under the curve from \(x=0\) to \(x=2\) is exactly 1 square unit. It's important to recognize that the 'area under the curve' refers to the actual geometric area without consideration of the sign. Whereas 'net area', as previously discussed, accounts for the layers of positivity and negativity created by the graph's relation to the x-axis. Mastering the distinction and calculation of these areas equips students with the tools to solve a wide array of problems within mathematics and applied sciences.