Problem 14 - Thomas Calculus

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 $$ (a)\int_{1}^{3} h(r) d r \quad \text { b. }-\int_{3}^{1} h(u) d u $$

Step-by-Step Solution

Answer

(a) 6; (b) -6.

Step 1: Interpret the given integrals

We know that \(\int_{-1}^{1} h(r) \, dr = 0\) and \(\int_{-1}^{3} h(r) \, dr = 6\). The first integral tells us that the function \(h\) accumulates zero net area between \(-1\) and \(1\). The second integral represents the net accumulation of \(h\) from \(-1\) to \(3\), which is equal to 6.

Step 2: Use properties of integrals to express \(\int_{1}^{3} h(r) \, dr\)

We know that: \[ \int_{-1}^{3} h(r) \, dr = \int_{-1}^{1} h(r) \, dr + \int_{1}^{3} h(r) \, dr \] Given that \(\int_{-1}^{1} h(r) \, dr = 0\), substitute these values into the equation: \[ 6 = 0 + \int_{1}^{3} h(r) \, dr \] Therefore, \(\int_{1}^{3} h(r) \, dr = 6\).

Step 3: Calculate \(-\int_{3}^{1} h(u) \, du\) by reversing limits of integration

The integral \(\int_{3}^{1} h(u) \, du\) is the same as \(-\int_{1}^{3} h(u) \, du\) by properties of definite integrals (since reversing the limits of integration changes the sign of the integral): \[ -\int_{3}^{1} h(u) \, du = \int_{1}^{3} h(u) \, du \] We already found \(\int_{1}^{3} h(u) \, du = 6\). Therefore, \(-\int_{3}^{1} h(u) \, du = -6\).

Key Concepts

Definite Integrals

Definite integrals are a fundamental concept in calculus that provide a way to calculate the area under a curve, or more precisely, the net accumulation of a function over a specific interval. If you imagine a function as a landscape, then the definite integral helps you find out how much land lies between certain points. For a function to be integrable on a given interval, it simply needs to be well-behaved enough for the calculus to work with it—no huge jumps or infinite spikes.
The notation \( \int_{a}^{b} f(x) \, dx \) represents the definite integral of the function \( f \) from \( a \) to \( b \), where:\

• \( a \) and \( b \) are the limits of integration.
• \( dx \) denotes the variable of integration.

You can think of \( \int_{-1}^{1} h(r) \, dr = 0 \) as saying that the area above the x-axis precisely cancels out the area below it between these points, leading to a net area of zero.

Properties of Integrals

Certain properties of integrals can simplify calculations and help us solve complex problems. Let's review the ones useful in this exercise:

• Linearity: Integrals have the property of linearity, which means you can add or subtract integrals as long as you operate on the same interval. Therefore, \( \int (f(x) + g(x)) \, dx = \int f(x) \, dx + \int g(x) \, dx \).
• Additivity: You can break up an integral at a point \( c \) within the limits: \( \int_{a}^{b} f(x) \, dx = \int_{a}^{c} f(x) \, dx + \int_{c}^{b} f(x) \, dx \). This is exactly what we did in our problem with \( \int_{-1}^{3} h(r) \, dr = \int_{-1}^{1} h(r) \, dr + \int_{1}^{3} h(r) \, dr \).

This understanding allowed us to determine that \( \int_{1}^{3} h(r) \, dr = 6 \). Using these properties can make handling larger intervals much more manageable.

Reversing Limits of Integration

Reversing the limits of integration is a useful trick that involves flipping the start and end points of an integral. When you reverse these limits, the sign of the integral changes. It is denoted as:\

• \( \int_{a}^{b} f(x) \, dx = -\int_{b}^{a} f(x) \, dx \).

In practical terms, if going from \( a \) to \( b \) results in a positive area, the reverse journey \( b \) to \( a \) will result in a negative area. This principle is what led us to find \( -\int_{3}^{1} h(u) \, du \) in the exercise. We performed the switch from \( \int_{3}^{1} \) to \( -\int_{1}^{3} \), achieving the operation by reversing the limits.
Such reversals are a key element in calculus when evaluating the sign-implications of an integral, especially in defining the bounds of accumulating or balancing quantities.