Problem 59 - Calculus for Biology and Medicine

Question

In Problems 59-63, verify each inequality without evaluating the integrals. $$ \int_{0}^{1} x d x \geq \int_{0}^{1} x^{2} d x $$

Step-by-Step Solution

Answer

The integral \( \int_{0}^{1} x \; dx \) is greater than or equal to \( \int_{0}^{1} x^2 \; dx \) because \( x \geq x^2 \) for \( x \) in [0,1].

Step 1: Understand the Inequality Problem

We are asked to verify the inequality \( \int_{0}^{1} x \; dx \geq \int_{0}^{1} x^{2} \; dx \). This means we need to show that the area under the curve \( y = x \) from 0 to 1 is greater than or equal to the area under the curve \( y = x^2 \) over the same interval.

Step 2: Analyze the Function Graphs

Consider the two functions: \( y = x \) and \( y = x^2 \). For any \( x \) in the interval [0,1], \( x \geq x^2 \), since \( x^2 - x = x(x - 1) \leq 0 \). This implies, graphically, that the line \( y = x \) is always above the parabola \( y = x^2 \) on this interval.

Step 3: Conclusion from Graph Analysis

Since the curve \( y = x \) is always above the curve \( y = x^2 \) from 0 to 1, the area under \( y = x \) (which is the integral \( \int_{0}^{1} x \; dx \)) is greater than the area under \( y = x^2 \) (which is \( \int_{0}^{1} x^{2} \; dx \)). Therefore, the inequality \( \int_{0}^{1} x \; dx \geq \int_{0}^{1} x^{2} \; dx \) holds.

Key Concepts

Understanding Integrals

Integrals are a fundamental concept in calculus that help us measure the area under a curve over a specific interval on a graph. In simpler terms, when we integrate a function, we are calculating the total area that lies between the function's curve and the x-axis. This area can be thought of as an accumulation of infinitesimally small segments. For example, for the function \( y = x \) over the interval from 0 to 1, the integral \( \int_{0}^{1} x \; dx \) gives us the area under this curve within those boundaries. Similarly, \( \int_{0}^{1} x^2 \; dx \) measures the area under the curve of \( y = x^2 \).
This calculation doesn't just tell us about the area but offers insights into the function's behavior over the interval. For comparing functions through their integrals, we look at their accumulated areas to determine which function's curve encloses more space within given limits.

Graph Analysis and Interpretation

Graph analysis is a key tool when comparing functions in the realm of inequalities. By visualizing the functions \( y = x \) and \( y = x^2 \) over the interval [0,1], we start to understand their relationship. The line represented by \( y = x \) is always above the parabola \( y = x^2 \) within this range. This relationship is crucial because it indicates that at each point from 0 to 1, the value of \( y = x \) is greater than or equal to \( y = x^2 \).
Here's how you can think about it:

• For any point \( x \) in the interval [0,1], the height of \( y = x \) is more than the height of \( y = x^2 \) because the parabola curves downwards more sharply compared to the straight line trajectory.
• Since the line is always above the parabola, it implies that over the same segment length, more area is captured by the line's curve compared to the parabola.

Thus, graphically analyzing these functions helps in verifying the given inequality without needing to compute actual integral values.

Comparing Functions using Inequalities

When comparing functions through inequalities, it's essential to determine how the functions behave relative to each other over the specified interval. The inequality \( \int_{0}^{1} x \; dx \geq \int_{0}^{1} x^2 \; dx \) tells us that the cumulative area under the curve of \( y = x \) is larger than that under \( y = x^2 \) from 0 to 1.
To understand this at a basic level:

• Each function essentially represents a type of growth. For \( y = x \), the line grows at a constant rate, while \( y = x^2 \) starts slower because it is squared and grows more rapidly after 1.
• In the interval from 0 to 1, since \( x \geq x^2 \), the linear function essentially "wins," capturing more area.

This use of inequalities in calculus allows us to conclude functional behavior without evaluating the integrals themselves. By understanding which function is larger over the particular bounds, we can draw conclusions about their respective integrals' total areas.