Problem 29

Question

29-34 Evaluate the integral by interpreting it in terms of areas. $$\int_{0}^{3}\left(\frac{1}{2} x-1\right) d x$$

Step-by-Step Solution

Verified

Answer

The integral evaluates to -0.75.

1Step 1: Identify the Function

Look at the integrand, which is $ \frac{1}{2}x - 1 $. This represents the function $ f(x) $ that we need to interpret in terms of area. The graph of $ f(x) = \frac{1}{2}x - 1 $ is a straight line.

2Step 2: Sketch the Graph of the Function

Draw a quick sketch of the line $ y = \frac{1}{2}x - 1 $. The line has a y-intercept at (0, -1) and a slope of $ \frac{1}{2} $, meaning it rises 1 unit for every 2 units it moves to the right.

3Step 3: Determine the Zeros (x-intercept)

To find where the line crosses the x-axis, set $ \frac{1}{2}x - 1 = 0 $. Solving for $ x $, we get $ \frac{1}{2}x = 1 $, so $ x = 2 $. This tells us the line crosses the x-axis at the point (2,0).

4Step 4: Divide the Area into Recognizable Shapes

Between $ x = 0 $ and $ x = 3 $, the line creates a right triangle below the x-axis from $ x = 0 $ to $ x = 2 $, and a smaller triangle above the x-axis from $ x = 2 $ to $ x = 3 $.

5Step 5: Calculate the Area Below the x-axis

The area of the triangle below the x-axis (from 0 to 2) can be calculated using the formula $ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} $. Here, the base is 2 (from $ x = 0 $ to $ x = 2 $) and the height is 1 (the y-intercept). Thus, the area is $ \frac{1}{2} \times 2 \times 1 = 1 $. This area is considered negative because it is below the x-axis.

6Step 6: Calculate the Area Above the x-axis

The area of the triangle above the x-axis (from 2 to 3) is found similarly. The base is 1 (from $ x = 2 $ to $ x = 3 $) and the height is 0.5 (from the slope $\frac{1}{2}$, calculated at $ x = 3 $: $ \frac{1}{2} \times 3 - 1 = \frac{1}{2} $). The area is $ \frac{1}{2} \times 1 \times 0.5 = 0.25 $. This area is positive because it is above the x-axis.

7Step 7: Find the Net Area

Add the areas computed: $-1 $ for the area below and $ 0.25 $ for the area above the x-axis. The net area is $-1 + 0.25 = -0.75$. Therefore, this is the value of the integral.

Key Concepts

Area under a curveDefinite integralGeometric interpretation of integrals

Area under a curve

The area under a curve refers to the space between the graph of a function and the x-axis. It's a concept that helps us understand integrals in a more geometric way. When we compute the integral of a function over an interval, we're essentially finding the total area trapped between the curve and the axis within those boundaries.

The area above the x-axis is positive because it aligns with the direction of positive y-values.
Conversely, the area below the x-axis is negative, which aligns with negative y-values.

To grasp this concept fully, imagine a landscape filled with hills (above the x-axis) and valleys (below the x-axis). The integral involves summing up the areas of both the hills and valleys. In the exercise, this was showcased with a linear function forming triangles above and below the axis, and calculating these provided the integral's value.

Definite integral

A definite integral is the integration of a function with fixed upper and lower boundaries. It gives us a precise numerical value representing the total area under a curve between two points (limits). In mathematical terms, if you see an integral with numbers as boundaries, it's a definite integral. Appreciate its utility in compaction:

Definite integrals help us find exact values quickly, like the total displacement or the area in physical spaces.
They simplify complex curve areas into manageable calculations.

The notation used in the exercise, such as $ \int_{0}^{3}\left(\frac{1}{2} x-1\right) d x$, indicates that the function $ \frac{1}{2} x - 1 $ is being evaluated from $ x = 0 $ to $ x = 3 $. This results in a value that truly represents the net area between this function and the x-axis within the specified interval.

Geometric interpretation of integrals

The geometric interpretation of integrals translates mathematical expressions into visual representations. This is particularly helpful in understanding how the integral gives a total sum of areas.

For the given problem: Picture the graph of $ y = \frac{1}{2}x - 1 $: it's a straight line.
From this line, calculate areas under the curve by identifying shapes it forms with axes, typically recognized as triangles, rectangles, or other geometric shapes.

In the process shown in the exercise, we broke the area into a triangle below the x-axis and another triangle above. Each triangle's area was calculated using basic geometric formulas. Subtracting the area below from the area above gave a better understanding of the integral's output, which was a total of $-0.75$. This approach not only aids in visual learning but also clarifies abstract integral calculus into something tangible and observable.

Problem 29

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