Problem 13 - 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. $$ \int_{2}^{1}[2 f(s)+5 g(s)] d s $$

Step-by-Step Solution

Answer

The value of the integral is \(-31\).

Step 1: Reverse the Limits of Integration

The integral is given as \( \int_{2}^{1} [2f(s) + 5g(s)] \, ds \). When reversing the limits of integration, we have to change the sign of the integral. Thus, \( \int_{2}^{1} [2f(s) + 5g(s)] \, ds = -\int_{1}^{2} [2f(s) + 5g(s)] \, ds \).

Step 2: Apply Linearity of Integrals

Using the property of linearity, split the integral into two separate integrals: \(-\int_{1}^{2} [2f(s) + 5g(s)] \, ds = -\left( 2\int_{1}^{2} f(s) \, ds + 5\int_{1}^{2} g(s) \, ds \right)\). This becomes \(-2\int_{1}^{2} f(s) \, ds - 5\int_{1}^{2} g(s) \, ds \).

Step 3: Solve \( \int_{1}^{2} g(s) \, ds \) Using Interval Additivity

We have \( \int_{0}^{2} g(s) \, ds = 4 \) and \( \int_{0}^{1} g(s) \, ds = -1 \). Using interval additivity, we find \( \int_{1}^{2} g(s) \, ds = \int_{0}^{2} g(s) \, ds - \int_{0}^{1} g(s) \, ds = 4 - (-1) = 5 \).

Step 4: Substitute Known Values into the Expression

We know from the problem \( \int_{1}^{2} f(s) \, ds = 3 \). Using this, substitute \( \int_{1}^{2} f(s) \, ds = 3 \) and \( \int_{1}^{2} g(s) \, ds = 5 \) into the expression from Step 2: \(-2(3) - 5(5)\).

Step 5: Simplify the Expression

Calculate \(-2(3) - 5(5) = -6 - 25\). This equals \(-31\).

Key Concepts

Properties of Integrals

Definite integrals have various properties that are essential in calculus problems. One of the core properties is the notion that integrals can be used to calculate the area under a curve between two points on the x-axis. These properties make integrals a powerful tool in analyzing functions.

• **Sign of limits**: Reversing the limits of integration reverses the sign of the integral. For example, given an integral from 2 to 1, this is equivalent to the negative of the integral from 1 to 2.
• **Additivity and splitting**: The definite integral of a function over a union of intervals is the sum of the integrals over each interval. This is particularly useful when we are given different intervals and need to compute an unknown part using known entries.

By understanding and applying these properties, such as reversing limits and additivity, you can solve calculus problems more effectively and with greater clarity.

Linearity of Integrals

The linearity of integrals is an essential property that allows us to simplify complex problems involving integrals. Linearity refers to two main properties:

• If you multiply a function by a constant and integrate, you can "factor out" the constant from the integral.
• Sums of functions can be integrated term by term. In other words, \[\int (af(x) + bg(x)) \, dx = a \int f(x) \, dx + b \int g(x) \, dx\]where \(a\) and \(b\) are constants, and \(f(x)\), \(g(x)\) are functions.

This property allows the integral \[- \int_{1}^{2} [2f(s) + 5g(s)] \, ds\] to be broken into manageable parts: \[-2 \int_{1}^{2} f(s) \, ds - 5 \int_{1}^{2} g(s) \, ds\].
This simplification is crucial in executing calculus problem-solving effectively.

Interval Additivity

Interval additivity is a powerful concept that helps in evaluating the integral of a function over an interval by using known integrals over sub-intervals. This property asserts that the definite integral over a whole interval can be split into the sum of integrals over subdivisions of that interval. For instance, if \[\int_{a}^{c} g(x) \, dx = K \] and we know \[\int_{a}^{b} g(x) \, dx \] then it follows that \[\int_{b}^{c} g(x) \, dx = K - \int_{a}^{b} g(x) \, dx\] provided \(a < b < c\). This concept was used to find \(\int_{1}^{2} g(s) \, ds\) by rearranging the relationships given in the original exercise. Understanding and employing interval additivity allows us to use information we already have to find unknown integrals.

Calculus Problem Solving

Calculus problem solving involving integrals often requires a multifaceted approach to break down and solve complex problems. When tackling a problem involving definite integrals, it is crucial to carefully apply properties such as linearity and interval additivity, as demonstrated in the original step by step solution.

• **Understand the problem**: Identify the given integrals and the limits of integration.
• **Use properties tactically**: Reverse limits when necessary, apply linearity to break into simpler integrals, and use additivity for complex intervals.
• **Simplify**: Substitute known values to progressively solve the problem.

In the example, understanding each of these steps combined with clear application of the integral properties leads to the final answer: \(-31\). Conquering calculus problems requires practice and careful attention to these properties and methods.