Problem 16 - Calculus Early Transcendentals: Pearson New International Edition

Question

Suppose that \(\int_{0}^{1} f(x) d x=2, \int_{1}^{2} f(x) d x=3, \int_{0}^{1} g(x) d x=-1\), and \(\int_{0}^{2} g(x) d x=4\). Use properties of definite integrals (linearity, interval additivity, and so on) to calculate each of the integrals in Problems 9-16. $$ \int_{0}^{2}[\sqrt{3} f(t)+\sqrt{2} g(t)+\pi] d t $$

Step-by-Step Solution

Answer

The integral evaluates to \(5\sqrt{3} + 4\sqrt{2} + 2\pi\).

Step 1: Apply Linearity of Integrals

The integral can be separated using the linearity of integrals:\[\int_{0}^{2} \left[\sqrt{3} f(t) + \sqrt{2} g(t) + \pi \right] dt = \sqrt{3} \int_{0}^{2} f(t) dt + \sqrt{2} \int_{0}^{2} g(t) dt + \int_{0}^{2} \pi \, dt\]

Step 2: Calculate Each Integral Separately

Let's handle each integral one at a time:1. **Integral of** \(f(t)\): \[ \int_{0}^{2} f(t) dt = \int_{0}^{1} f(t) dt + \int_{1}^{2} f(t) dt = 2 + 3 = 5 \]2. **Integral of** \(g(t)\): Given, \(\int_{0}^{2} g(t) dt = 4\).3. **Integral of** \(\pi\): \[ \int_{0}^{2} \pi \, dt = \pi \times (2 - 0) = 2\pi \]

Step 3: Combine Results

Now substitute the values back into the separated integral:\[\sqrt{3} \times 5 + \sqrt{2} \times 4 + 2\pi\]This simplifies to:\[5\sqrt{3} + 4\sqrt{2} + 2\pi\]

Key Concepts

Linearity of Integrals

The linearity of integrals is a fundamental property that simplifies the process of integrating functions. It tells us that the integral of a sum of functions is the sum of their integrals. More formally, for any constants \(a\) and \(b\), and functions \(f(x)\) and \(g(x)\), we have:

• \(\int [a f(x) + b g(x)] \, dx = a \int f(x) \, dx + b \int g(x) \, dx\)

This property is incredibly useful because it allows us to break down complex expressions into manageable parts. In our given exercise, it enables us to separate the whole expression \(\int_{0}^{2} [\sqrt{3} f(t) + \sqrt{2} g(t) + \pi] \, dt\) into three distinct integrals. This separation makes the calculation straightforward, as we can simply integrate each part individually:

• \(\sqrt{3} \int_{0}^{2} f(t) \, dt\)
• \(\sqrt{2} \int_{0}^{2} g(t) \, dt\)
• \(\int_{0}^{2} \pi \, dt\)

By applying the linearity property, we streamline the process, ensuring accuracy and clarity in solving the integral.

Integration Techniques

Integration techniques are strategies that help us compute the integral of complex functions. In our problem, we're using straightforward techniques aided by given information about specific integrals. Let's discuss them briefly:

• Direct Substitution: We directly use the given integrals, such as \(\int_{0}^{2} g(t) \, dt = 4\), to plug into our final expression with no need for further manipulation.
• Scalar Multiplication: Whenever constants, like \(\sqrt{3}\) and \(\sqrt{2}\), multiply the functions within the integrals, we can factor them out, simplifying our calculations. This allows us to focus on determining \(\int_{0}^{2} f(t) \, dt\) separately, using known values.

These techniques are efficient because they leverage known values and mathematical properties to solve definite integrals quickly. Understanding when and how to apply these methods is crucial for solving complex integrals without unnecessary complexity.

Interval Additivity

Interval additivity is another key property of definite integrals that is extremely useful in solving problems. It says if you have an integral over an interval that can be split into subintervals, the integral over the entire interval is the sum of the integrals over the subintervals. Mathematically, for a function \(f(x)\) over \([a, c]\), if \(b\) is between \(a\) and \(c\):

• \(\int_{a}^{c} f(x) \, dx = \int_{a}^{b} f(x) \, dx + \int_{b}^{c} f(x) \, dx\)

Using this in our solution, we were able to compute \(\int_{0}^{2} f(t) \, dt\) by adding \(\int_{0}^{1} f(t) \, dt = 2\) and \(\int_{1}^{2} f(t) \, dt = 3\), resulting in 5. Hence, interval additivity allows us to decompose integrals across subintervals, facilitating the computation process by utilizing simpler intervals.