Problem 16 - Calculus for Biology and Medicine

Question

Assume that \(a<0

Step-by-Step Solution

Answer

The geometric argument shows \( \int_{a}^{b} x \, dx = \frac{b^{2} - a^{2}}{2} \).

Step 1: Understanding the Problem

We are given the problem of finding the definite integral \( \int_{a}^{b} x \, dx \) where \( a < 0 < b \). To solve this, we will use geometry to derive the expression \( \frac{b^{2} - a^{2}}{2} \).

Step 2: Geometric Representation

The integral \( \int_{a}^{b} x \, dx \) represents the area under the line \( y = x \) from \( x = a \) to \( x = b \). The graph of \( y = x \) is a diagonal line through the origin. The area under this line from \( a \) to \( b \) can be thought of as the areas of two triangles: one from \( a \) to 0, and another from 0 to \( b \).

Step 3: Calculating Areas of Triangles

The area of a triangle is \( \frac{1}{2} \times \text{base} \times \text{height} \). - For the triangle from \( a \) to 0, the base is \( |a| \), and the height is \( |a| \), so the area is \( \frac{1}{2} a^{2} \).- For the triangle from 0 to \( b \), the base is \( b \), and the height is \( b \), providing an area of \( \frac{1}{2} b^{2} \).

Step 4: Net Area and the Expression

Since \( a < 0 \) and \( b > 0 \), the area from \( a \) to 0 is below the x-axis and subtracted from the area from 0 to \( b \). Thus, the total area is \( \frac{1}{2} b^{2} - \left(-\frac{1}{2} a^{2}\right) = \frac{b^{2} + a^{2}}{2} \), which simplifies to \( \frac{b^{2} - a^{2}}{2} \).

Step 5: Expressing Limits as Definite Integrals

To express any subinterval as a definite integral, observe that each subinterval \([x_i, x_{i+1}]\) with width \( w = \frac{b-a}{n} \) can be expressed as \( \int_{x_i}^{x_{i+1}} \) for a function over that subinterval. However, we are only given the integral from \( a \) to \( b \), already expressed in this form.

Key Concepts

Geometric Interpretation

Understanding the definite integral \( \int_{a}^{b} x \, dx \) in a geometric sense helps us to visualize what the calculation represents. When you have the graph of the function \( y = x \), it creates a diagonal line through the origin with a slope of 1. This line, stretching from \( x = a \) to \( x = b \), forms a total area under the curve within these limits.

The integral essentially quantifies the net area captured between this line and the x-axis. Since \( a < 0 < b \), we handle two main regions:

• The region from \( a \) to 0, where the area lies below the x-axis.
• The region from 0 to \( b \), where the area is above the x-axis.

Both these regions can be seen as forming right triangles along the y-axis. Hence, the geometric interpretation simplifies the integral as the total area of these two triangles. The problem is elegantly boiled down to a matter of calculating these straight-edged shapes, making the integral more tangible and understandable by connecting algebra with geometry.

Area Under Curve

The fundamental task of determining \( \int_{a}^{b} x \, dx \) is calculating the area under the curve of the function \( y = x \) between the specified points. But instead of dealing with any complex function, here it's merely identifying the areas of geometric shapes — specifically triangles.

When \( a < 0 \) and \( b > 0 \), the graph of \( y = x \) shows two distinct triangular areas:

• From \( a \) to 0: Since this part of the graph is under the x-axis, the area can be calculated using the triangle formula \( \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \). It results in \( \frac{1}{2} a^{2} \).
• From 0 to \( b \): This area is above the x-axis, leading to a positive contribution, calculated similarly as \( \frac{1}{2} b^{2} \).

To find the net area, subtract the negative area (from \( a \) to 0) from the positive area (from 0 to \( b \)). Thus, the result becomes \( \frac{b^{2} - a^{2}}{2} \), the exact value of the original integral.

Triangles Under Line

Triangles form a fundamental way to interpret the integral \( \int_{a}^{b} x \, dx \) when viewed through the lens of geometry. By analyzing the line \( y = x \), we identify two right triangles reflecting the areas we consider:

• The triangle from \( a \) to 0 has both its base and its height equivalent to \( |a| \). This is because it spans across both the negative x-axis and the line \( y = x \).
• The triangle from 0 to \( b \) has its base and height equal to \( b \). This is a result of the entirely positive stretch along the axis.

The geometric properties of these triangles make the integral calculation straightforward. Once we calculate each triangle's area by \( \frac{1}{2} \times \text{base} \times \text{height} \), we combine them adjusting for the area below the axis (negative) and above (positive), thus leading us to the nicely simplified form: \( \frac{b^{2} - a^{2}}{2} \). This approach bridges understanding from simple geometry to more complex calculus, showcasing the powerful synergy of math's different branches.