Problem 14 - University Calculus: Early Transcendentals

Question

Suppose that \(h\) is integrable and that \(\int_{-1}^{1} h(r) d r=0\) and \(\int_{-1}^{3} h(r) d r=6 .\) Find A. \(\int_{1}^{3} h(r) d r\) B. \(-\int_{3}^{1} h(u) d u\)

Step-by-Step Solution

Answer

A: 6, B: 6.

Step 1: Understanding the Problem

We need to find two integration values based on the given integrals of the integrable function \( h(r) \). We have two pieces of information: \( \int_{-1}^{1} h(r) \, dr = 0 \) and \( \int_{-1}^{3} h(r) \, dr = 6 \). Using these, we can determine the other requested integrals.

Step 1: Calculating \( \int_{1}^{3} h(r) \, dr \)

First, we use the property of definite integrals which states that \( \int_{a}^{c} f(x)\,dx = \int_{a}^{b} f(x)\,dx + \int_{b}^{c} f(x)\,dx \). In our problem, this gives us the equation\[\int_{-1}^{3} h(r) \, dr = \int_{-1}^{1} h(r) \, dr + \int_{1}^{3} h(r) \, dr.\]Substitute the known values:\[6 = 0 + \int_{1}^{3} h(r) \, dr.\]Thus, \( \int_{1}^{3} h(r) \, dr = 6 \).

Step 2: Calculating \( -\int_{3}^{1} h(u) \, du \)

Recalling the property of definite integrals known as mutual reversal, \( \int_{a}^{b} f(x) \, dx = -\int_{b}^{a} f(x) \, dx \), we have:\[-\int_{3}^{1} h(u) \, du = \int_{1}^{3} h(u) \, du.\]Since we have already computed \( \int_{1}^{3} h(r) \, dr = 6 \), it follows that \[-\int_{3}^{1} h(u) \, du = 6.\]

Key Concepts

Integral Properties

Integral properties are essential tools that help us solve and understand definite integrals. These properties simplify computations and provide deeper insights into the behavior of functions. Let’s delve into some crucial properties that were used in the problem:

• Linearity of integrals: This property states that the integral of a sum is the sum of the integrals. This allows us to break integrals into more manageable parts. For a function \( h(r) \), it uses the equation \[ \int_{a}^{c} h(r) \, dr = \int_{a}^{b} h(r) \, dr + \int_{b}^{c} h(r) \, dr \] as seen in the solution process.


• Additivity of intervals: Within linearity, the additivity of intervals allows us to evaluate an integral over a composite interval by adding integrals over sub-intervals. This was specifically used in the problem to split the integral from \(-1\) to \(3\) into two: from \(-1\) to \(1\) and from \(1\) to \(3\).

Understanding these integral properties makes the evaluation of definite integrals more straightforward and comprehensible. They serve as the foundation for breaking down complex problems into easy-to-solve parts.

Mutual Reversal

The concept of mutual reversal in definite integrals reveals an interesting symmetry property. Mutual reversal states that reversing the limits of an integral changes the sign of the result:

• The integral \( \int_{a}^{b} h(x) \, dx \) is equal to \(-\int_{b}^{a} h(x) \, dx \).

This property is a direct consequence of how areas under curves are considered positive or negative based on direction. In the original exercise, mutual reversal was applied to find the integral from 3 to 1 by reversing it to go from 1 to 3. As a result, this manipulation simplified finding \(-\int_{3}^{1} h(u) \, du\), avoiding recalculation and using existing knowledge. It showcases how reversing limits shifts our perspective but does not change the absolute value of the integral.

Integrable Function

An integrable function is one for which a definite integral can be evaluated over a given interval. In simpler terms, it's a function that doesn't "misbehave" too much. For a function to be integrable on an interval, certain conditions must be met:

• The function should be bounded on the interval, meaning it doesn't go to infinity.


• There should be a finite number of discontinuities. Typically, it can still be integrated if these discontinuities are jump discontinuities.

Understanding that \( h(r) \) is integrable is crucial because it assures that the definite integrals used in the problem can be reliably calculated. Even though no specific details about \( h(r) \) are given, knowing it is integrable means we can apply integral formulas and properties confidently. This ensures the results derived from these operations hold true in a mathematical sense.