Problem 19 - The Calculus with Analytic Geometry

Question

Show that \(\int_{0}^{1} x d x \geq \int_{0}^{1} x^{2} d x\) but \(\int_{1}^{2} x d x \leq \int_{1}^{2} x^{2} d x\). Do not evaluate the definite integrals.

Step-by-Step Solution

Answer

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

Step 1 - Compare the Integrands

Examine the functions inside the integrals, which are \(x\) and \(x^2\). For \(0 \leq x \leq 1\), note that \(x \geq x^2\). Conversely, for \(1 \leq x \leq 2\), note that \(x \leq x^2\).

Step 2 - Analyze the Interval from 0 to 1

Since \(x \geq x^2\) in the interval \([0, 1]\), the integral of \(x\) over \([0, 1]\) is greater than or equal to the integral of \(x^2\) over \([0, 1]\).

Step 3 - Analyze the Interval from 1 to 2

Since \(x \leq x^2\) in the interval \([1, 2]\), the integral of \(x\) over \([1, 2]\) is less than or equal to the integral of \(x^2\) over \([1, 2]\).

Key Concepts

Integral Inequalities

An integral inequality shows the relationship between the definite integrals of two functions over a certain interval. If one function is greater than another function over an interval, the integral of the first function will also be greater over that interval.
Specifically, for functions \(f(x)\) and \(g(x)\):

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

This principle allows us to understand how integrals compare without calculating their exact values.

Function Comparison

When comparing functions in the context of integrals, it’s crucial to determine their relationship over the specified interval.
For instance, let’s consider two functions \(f(x) = x\) and \(g(x) = x^2\). To compare them, we analyze their values over the given intervals:

• For the interval \([0, 1]\): Here, \(f(x) = x\) is greater than \(g(x) = x^2\). This is because \(x\) is directly proportional to \(x\), while \(x^2\) grows faster but starts smaller. Thus, \(f(x) \geq g(x)\).
• For the interval \([1, 2]\): In this case, \(g(x) = x^2\) is greater than or equal to \(f(x) = x\) as \(x^2\) increases more rapidly than \(x\). Thus, \(f(x) \leq g(x)\).

Understanding this relationship helps in setting up the integral comparison without actual computation.

Integration Intervals

The intervals over which integration is performed play a critical role in determining the relationship between integrals of different functions.
In definite integrals, we work within limits that define a starting and ending point for the area under the curve.
For our given problem, the intervals are \([0, 1]\) and \([1, 2]\):

• On \([0, 1]\): Consider functions \(x\) and \(x^2\). Since \(x \geq x^2\), their integrals will also respect this inequality. Thus, \( \int_{0}^{1} x \, dx \geq \int_{0}^{1} x^2 \, dx\).
• On \([1, 2]\): Here, we have the reverse relationship. As \(x^2\) grows faster, \(x \leq x^2\). Hence, \( \int_{1}^{2} x \, dx \leq \int_{1}^{2} x^2 \, dx\).

This analysis simplifies the process by focusing on interval properties and relationships.