Problem 7 - Calculus Single Variable

Question

Suppose that \(\int_{9}^{4} f(x) d x=5\) and \(\int_{9}^{4} g(x) d x=15 .\) Evaluate \(\int_{4}^{9}(6 f(x)-7 g(x)) d x\).

Step-by-Step Solution

Answer

The integral evaluates to 75.

Step 1: Understand Properties of Definite Integrals

The properties of definite integrals state that you can split and factor integrals respectively. This means \( \int_{a}^{b} [ cf(x) + dg(x) ] \, dx = c \int_{a}^{b} f(x) \, dx + d \int_{a}^{b} g(x) \, dx \) where \(c\) and \(d\) are constants.

Step 2: Reverse Limits Property

The property \( \int_{a}^{b} f(x) \, dx = -\int_{b}^{a} f(x) \, dx \) allows us to reverse the limits of the given integrals. Thus, \( \int_{4}^{9} f(x) \, dx = -\int_{9}^{4} f(x) \, dx = -5 \) and \( \int_{4}^{9} g(x) \, dx = -\int_{9}^{4} g(x) \, dx = -15 \).

Step 3: Substitute into the Function

Substitute the values we derived into the integral \( \int_{4}^{9}(6 f(x)-7 g(x)) \, dx \). This becomes: \( 6 \int_{4}^{9} f(x) \, dx - 7 \int_{4}^{9} g(x) \, dx \).

Step 4: Calculate Using Properties

Calculate each part: \( 6 \times (-5) = -30 \) and \( -7 \times (-15) = 105 \).

Step 5: Combine Results

Add both parts: \( -30 + 105 = 75 \). So, \( \int_{4}^{9}(6 f(x)-7 g(x)) \, dx = 75 \).

Key Concepts

Properties of Definite Integrals

Understanding the properties of definite integrals is crucial when dealing with integrals involving more than one function. These properties allow you to break down complex integrals into simpler parts, making them easier to handle and evaluate. Let's explore these properties:

• Linear Property: This states that if you have a linear combination of functions, such as \( c f(x) + d g(x) \), the integral over an interval can be split. This means you can write it as \( c \int_{a}^{b} f(x) \, dx + d \int_{a}^{b} g(x) \, dx \).
• Constant Multiplication: If a constant multiplies a function inside an integral, you can factor the constant out of the integral. For example, \( c \int_{a}^{b} f(x) \, dx = \int_{a}^{b} c f(x) \, dx \).

These properties simplify evaluating more complicated expressions by allowing you to manipulate integrals algebraically before diving into more detailed computations.

Reverse Limits Property

One of the interesting properties of definite integrals is the ability to reverse the limits of integration. This can be quite useful in different scenarios, particularly when the given integral needs to be manipulated for further evaluation. Let's dive into this property:

• Reverse Limits Formula: The property \( \int_{a}^{b} f(x) \, dx = -\int_{b}^{a} f(x) \, dx \) signifies that by reversing the limits, you effectively change the sign of the integral. This is because the integral measures the signed area under the curve.
• Application Example: In the problem at hand, if you initially have \( \int_{9}^{4} f(x) \, dx \), and you wish to evaluate \( \int_{4}^{9} f(x) \, dx \) instead, you simply flip the integration limits and adjust the sign. Thus, \( \int_{4}^{9} f(x) \, dx = -5 \) when \( \int_{9}^{4} f(x) \, dx = 5 \).

This swap in limits is handy and frequently used especially when combining multiple integrals as it helps match their intervals.

Evaluating Integrals

Once you understand the properties and manipulations of definite integrals, the next step is evaluating them. Here's how you can evaluate integrals such as the one given in the exercise:

• Use Derived Values: Based on the initial properties, compute individual components of the integral separately. For example, if \( \int_{4}^{9} f(x) \, dx = -5 \) and \( \int_{4}^{9} g(x) \, dx = -15 \), these values are fundamental to further calculations.
• Substitute and Simplify: Substitute these derived values into any new combined integrals. For instance, \( 6 \int_{4}^{9} f(x) \, dx - 7 \int_{4}^{9} g(x) \, dx \) becomes \( 6 \times (-5) - 7 \times (-15) \).
• Compute Result: Execute the arithmetic calculations to find the final result. In our example, \( 6 \times (-5) = -30 \) and \( -7 \times (-15) = 105 \). Adding these gives you the final result, \( -30 + 105 = 75 \).

Applying these methods lets you evaluate even complex integrals systematically by simplifying through properties and performing precise arithmetic.