Problem 75 - Calculus for Biology and Medicine

Question

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

\(\int_{0}^{1} x \,dx \geq \int_{0}^{1} x^2 \,dx\) because for all \(x \in [0, 1], x \geq x^2\).

Step 1: Analyze the Functions

First, we analyze the functions within the integrals: \(f(x) = x\) and \(g(x) = x^2\). Notice that for all \(x\) in the interval \([0, 1]\), \(f(x) = x\) is greater than or equal to \(g(x) = x^2\), because the function \(x\) increases linearly while \(x^2\) increases quadratically and thus is always less than or equal to \(x\) within this range.

Step 2: Use of Integral Comparison Property

According to the integral comparison property, if \(f(x) \geq g(x)\) on an interval \([a, b]\), then \(\int_{a}^{b} f(x) \,dx \geq \int_{a}^{b} g(x) \,dx\). Since we showed in Step 1 that \(x \geq x^2\) for \(x \in [0, 1]\), it follows directly that \(\int_{0}^{1} x \,dx \geq \int_{0}^{1} x^2 \,dx\).

Key Concepts

Integral Comparison Property

The integral comparison property is a powerful tool when you want to compare the size of integrals without directly evaluating them. This property states that if you have two functions, say \( f(x) \) and \( g(x) \), and one function is always greater than or equal to the other over a certain interval \([a, b]\), then the integral of the greater function over that interval will also be greater, or at least equal to, the integral of the lesser function.

Suppose you have:

• \( f(x) \geq g(x) \) for all \( x \) in \([a, b]\)
• Then, \( \int_{a}^{b} f(x) \, dx \geq \int_{a}^{b} g(x) \, dx \)

This makes it especially useful for verifying inequalities like the one presented in our original exercise. It allows you to conclude the relationship of the integrals based on the relationship of the functions alone. No need to complicatedly evaluate the integrals if this condition holds true.

By employing this property, you save time and effort, efficiently drawing conclusions from inequalities between functions themselves.

Functions Analysis

Examining the behavior of functions is a fundamental step in understanding their relationships over a given interval. In the original exercise, we compared the functions \( f(x) = x \) and \( g(x) = x^2 \).

The way these functions grow is key.
- **Linear Function:** \( x \) is linear; it steadily increases and has a constant rate of change.- **Quadratic Function:** \( x^2 \) is quadratic; it grows more slowly than \( x \) initially but then accelerates rapidly as \( x \) increases beyond 1.
However, between 0 and 1, \( x^2 \) will stay below \( x \) because it starts with large growth rates only after this interval.

While \( x^2 = x \) when \( x=1 \), in every other point of \([0, 1)\), \( x^2 \leq x \). This analysis confirms that \( f(x) \geq g(x) \) in this specific context, allowing us to utilize the integral comparison property confidently.

Inequality Verification

Verifying inequality between integrals without solving them involves some strategic thinking. You're primarily comparing the functions involved over a specified interval.

To verify that the inequality \( \int_{0}^{1} x \, dx \geq \int_{0}^{1} x^2 \, dx \) holds, start by looking at how each function behaves between 0 and 1, as covered in the functions analysis.

Once you've established that \( f(x) = x \) is always equal to or greater than \( g(x) = x^2 \) within this range, you apply the integral comparison property.

• Since \( x \geq x^2 \) for \( x \) in \([0, 1]\), their respective integrals will maintain this inequality: \( \int_{0}^{1} x \, dx \geq \int_{0}^{1} x^2 \, dx \).

There's no need for further complex calculations.
This logical approach saves time and reinforces your understanding of how integrals can be compared using inequalities.