Problem 23 - University Calculus: Early Transcendentals

Question

Use known area formulas to evaluate the integrals in Exercises \(23-28\) $$\int_{0}^{b} \frac{x}{2} d x, \quad b>0$$

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{1}{4}b^2 \).

Step 1: Identify the Geometric Shape

The integrand \( \frac{x}{2} \) forms a straight line when plotted against \( x \). Therefore, the area under the curve from 0 to \( b \) forms a right triangle above the \( x \)-axis.

Step 2: Find the Base and Height of the Triangle

The base of the triangle extends from \( 0 \) to \( b \), making it \( b \) units long. The height of the triangle is the value of the function at \( x = b \), which is \( \frac{b}{2} \).

Step 3: Apply the Area Formula of a Triangle

The area of a triangle is given by \( \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \). Substitute the base and height of our triangle into this formula: \( \text{Area} = \frac{1}{2} \times b \times \frac{b}{2} \).

Step 4: Simplify the Expression

Calculate the area: \( \text{Area} = \frac{1}{2} \times b \times \frac{b}{2} = \frac{1}{4}b^2 \). This area represents the integral of \( \frac{x}{2} \) from 0 to \( b \).

Key Concepts

Area under a Curve

The concept of "area under a curve" is fundamental in calculus. It involves finding the total area between a curve, represented by a function, and the x-axis over a specified interval. For a curve defined by a function \( f(x) \), the area is the space trapped between the x-axis and the curve from one point to another.

This area can be visually understood by considering specific geometric shapes enclosed by the curve. By analyzing these shapes, like rectangles or triangles, one may apply well-known geometric area formulas.

• This method helps in situations where the shapes can be recognized, simplifying the integration process.
• Converting the integration problem into a geometric problem is often easier when dealing with simple shapes.

The process essentially connects calculus with basic geometry, making the concept more approachable.

Definite Integrals

Definite integrals are a principal concept used to calculate the area under a curve over an interval. When you integrate a function \( \int_{a}^{b} f(x) \) over \([a,b]\), you are finding the accumulated area between the function \( f(x) \) and the x-axis from point \( a \) to point \( b \).

This is not just any area; it considers the direction relative to the axis:

• Areas above the x-axis count as positive.
• Areas below count as negative.

This characteristic makes definite integrals powerful, as they fully capture the behavior of a function over its interval. For practical calculation, we can use the geometry. In the example, the function forms a triangle simplifying integration to calculating the triangle's area: \(\frac{1}{2} \times \text{base} \times \text{height}\). Thus, recognizing and using definite integrals goes beyond simple arithmetic, revealing the depth of what integration can offer.

Geometric Interpretation of Integrals

The geometric interpretation of integrals invites us to see calculus as an extension of geometry, where the function \( f(x) \) sketches an area that can be calculated geometrically. When grappling with integration, visualizing the graph of the function can assist greatly.

• For simple functions, mapping them into recognizable shapes simplifies understanding the calculation involved in taking integrals.
• For instance, when a linear function like \( \frac{x}{2} \) is plotted, it creates a line that can form a familiar geometric shape, such as a triangle.

This is when basic formulas, like those for triangles or rectangles, become relevant.

By matching the areas under curves to known shapes, you can convert complex integral problems into straightforward geometric exercises. This approach to calculus provides intuition and a solid foundation for tackling more complex integrations.