Problem 13 - University Calculus: Early Transcendentals

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. \(\int_{3}^{4} f(z) d z\) b. \(\int_{4}^{3} f(t) d t\)

Step-by-Step Solution

Answer

a. 4, b. -4

Step 1: Understanding the Problem

We are given two definite integrals: \( \int_{0}^{3} f(z) \, dz = 3 \) and \( \int_{0}^{4} f(z) \, dz = 7 \). We need to use these to find a. \( \int_{3}^{4} f(z) \, dz \) and b. \( \int_{4}^{3} f(t) \, dt \).

Step 2: Applying the Fundamental Theorem of Calculus for Part a

To find \( \int_{3}^{4} f(z) \, dz \), use the property \( \int_{a}^{b} f(x) \, dx = \int_{0}^{b} f(x) \, dx - \int_{0}^{a} f(x) \, dx \). Substitute: \( \int_{3}^{4} f(z) \, dz = \int_{0}^{4} f(z) \, dz - \int_{0}^{3} f(z) \, dz \).

Step 3: Calculating the Result for Part a

Substitute the given integrals into the expression: \( \int_{3}^{4} f(z) \, dz = 7 - 3 \). Thus, \( \int_{3}^{4} f(z) \, dz = 4 \).

Step 4: Understanding the Integral from b

\( \int_{4}^{3} f(t) \, dt \) is simply the negative of \( \int_{3}^{4} f(t) \, dt \) due to the property \( \int_{a}^{b} f(x) \, dx = -\int_{b}^{a} f(x) \, dx \).

Step 5: Calculating the Result for Part b

From Step 3, we know that \( \int_{3}^{4} f(z) \, dz = 4 \). Hence, \( \int_{4}^{3} f(t) \, dt = -4 \).

Key Concepts

Definite Integrals

A definite integral provides a way to calculate the net area under a curve within a specified interval. When we say a function is integrable over an interval, it means we can find its integral, or area, over that range.

In the given exercise, the definite integral \( \int_{0}^{3} f(z) \, dz = 3 \) tells us that the net area under the curve \( f(z) \) from \( z = 0 \) to \( z = 3 \) is 3 square units. Similarly, \( \int_{0}^{4} f(z) \, dz = 7 \) tells us the net area under the curve from \( z = 0 \) to \( z = 4 \) is 7 square units.

To solve problems involving definite integrals, follow these steps:

• Identify the interval of integration.
• Apply any known properties or theorems, such as the Fundamental Theorem of Calculus.
• Use the given information to solve for the unknown integral.

Understanding definite integrals allows us to quantify the total accumulation of a function’s values over a specific interval.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus connects differentiation and integration, and it consists of two parts. Here, we focus on how it enables us to evaluate definite integrals.

One of the key properties expressed via the Fundamental Theorem is that if \( F \) is an antiderivative of \( f \) over an interval \( [a, b] \), then:
\[\int_{a}^{b} f(x) \, dx = F(b) - F(a)\]
In simpler terms, it allows us to find the value of a definite integral by evaluating the antiderivative at the boundaries \( a \) and \( b \). In our problem, we cleverly use this theorem as a property:

• We subtracted \( \int_{0}^{3} f(z) \, dz \) from \( \int_{0}^{4} f(z) \, dz \) to find \( \int_{3}^{4} f(z) \, dz \).

This theorem is a powerful tool because it simplifies the computation of integrals, especially when exact antiderivatives can be found.

Integral Properties

Several properties of integrals are used to solve complex problems efficiently. These properties help manipulate the limits and the integrand for more manageable calculations.

**Property 1: Additive Nature**
This property states that combining integral intervals is valid. For example, if we have:\[\int_{a}^{b} f(x) \, dx + \int_{b}^{c} f(x) \, dx = \int_{a}^{c} f(x) \, dx\]This helped us break down the region of integration in parts a and b of our exercise.

**Property 2: Reversal of Limits**

• Changing the order of limits reverses the sign of the integral:\[\int_{a}^{b} f(x) \, dx = -\int_{b}^{a} f(x) \, dx\]
• This was essential in our problem, showing that \( \int_{4}^{3} f(t) \, dt = -\int_{3}^{4} f(t) \, dt \).

By applying these properties, especially the reversal of limits, we were able to find the solution to part b. Knowing these properties enriches our toolbox for tackling different integral-related challenges in calculus.