Problem 13 - Thomas Calculus

Question

Suppose that \(f\) is integrable and that \(\int_{0}^{3} f(z) d z=3\) and \(\int_{0}^{4} f(z) d z=7 .\) Find $$ (a)\int_{3}^{4} f(z) d z \quad \text { b. } \int_{4}^{3} f(t) d t $$

Step-by-Step Solution

Answer

(a) 4, (b) -4

Step 1: Understanding Integral Properties

The integral \(\int_{a}^{b} f(z) dz \) represents the area under the curve of a function \(f(z)\) from \(z=a\) to \(z=b\). An essential property of definite integrals is that \( \int_{a}^{b} f(z) dz + \int_{b}^{c} f(z) dz = \int_{a}^{c} f(z) dz \). We'll use this property to find the unknown integrals.

Step 2: Calculating \\( \int_{3}^{4} f(z) dz \\) Using Known Values

From the properties of integrals, we have \(\int_{0}^{4} f(z) dz = \int_{0}^{3} f(z) dz + \int_{3}^{4} f(z) dz \). We know \(\int_{0}^{4} f(z) dz = 7\) and \(\int_{0}^{3} f(z) dz = 3\). Solving for \(\int_{3}^{4} f(z) dz \), we get \(7 = 3 + \int_{3}^{4} f(z) dz \), thus \(\int_{3}^{4} f(z) dz = 4\).

Step 3: Determining \\( \int_{4}^{3} f(t) dt \\)

The sign of the definite integral can be reversed by swapping the limits of integration, which implies that \(\int_{a}^{b} f(z) dz = -\int_{b}^{a} f(z) dz \). Therefore, \(\int_{4}^{3} f(t) dt = -\int_{3}^{4} f(t) dt \). Since we calculated that \(\int_{3}^{4} f(z) dz = 4\), it follows that \(\int_{4}^{3} f(t) dt = -4\).

Key Concepts

Integral Properties

When dealing with definite integrals, understanding their properties is crucial. Definite integrals are not just numbers; they represent the accumulated total, or area, under the curve of a function between specific limits. One fundamental property is the **additivity of integrals**. This means that if you know the integrals from point A to C (\( \int_{a}^{c} f(z) \, dz \)) and from A to B (\( \int_{a}^{b} f(z) \, dz \)), you can find the integral from B to C by subtraction:

• \( \int_{a}^{c} f(z) \, dz = \int_{a}^{b} f(z) \, dz + \int_{b}^{c} f(z) \, dz \)

This property makes it possible to solve for unknown integrals between specific segments when part of the information is missing. For instance, if you know \( \int_{0}^{4} f(z) \, dz = 7 \) and \( \int_{0}^{3} f(z) \, dz = 3 \), you can determine \( \int_{3}^{4} f(z) \, dz \) by rearranging them.

Area Under Curve

The concept of the area under the curve is a visual way of understanding what a definite integral represents. Imagine a graph of a function, the definite integral \( \int_{a}^{b} f(z) \, dz \) measures the total area between the curve of the function \( f(z) \) and the z-axis, from \( z=a \) to \( z=b \). This area can represent anything from physical quantities like distance or volume, depending on the function you're examining.

• The areas calculated using integrals can be positive if the function is above the x-axis, and negative if below.
• The total area, therefore, accounts for both positive and negative contributions, making the definite integral incredibly versatile.

Understanding the area under the curve allows us to solve real-world problems by computing values that represent physical quantities. This highlights why definite integrals are a powerful mathematical tool.

Reversing Limits of Integration

An interesting property of definite integrals is how their value changes when you reverse their limits. When you switch the limits of integration, the sign of the integral changes. Mathematically, this can be expressed as:

• \( \int_{a}^{b} f(z) \, dz = -\int_{b}^{a} f(z) \, dz \)

This property is particularly handy when evaluating integrals where limit reversal simplifies the calculation. For example, if \( \int_{3}^{4} f(z) \, dz = 4 \), reversing these limits gives \( \int_{4}^{3} f(z) \, dz = -4 \). This simple observation can save time and effort during problem-solving, thus making this property both fascinating and practically beneficial.