Problem 93 - Calculus Volume 1

Question

Suppose that \(\int_{0}^{4} f(x) d x=5\) and \(\int_{0}^{2} f(x) d x=-3,\) and \(\int_{0}^{4} g(x) d x=-1\) and \(\int_{0}^{2} g(x) d x=2 .\) In the following exercises, compute the integrals. \(\int_{2}^{4}(4 f(x)-3 g(x)) d x\)

Step-by-Step Solution

Answer

41

Step 1: Understand the Properties of Definite Integrals

We start by recognizing that we need to find \( \int_{2}^{4} (4f(x) - 3g(x)) \, dx \). We can use the property of linearity of definite integrals: if \( \int_a^b u(x) \, dx = U \) and \( \int_a^b v(x) \, dx = V \), then \( \int_a^b (cu(x) + dv(x)) \, dx = cU + dV \).

Step 2: Split the Integral Using Properties

Split the given integral using linearity: \[ \int_{2}^{4} (4f(x) - 3g(x)) \, dx = 4 \int_{2}^{4} f(x) \, dx - 3 \int_{2}^{4} g(x) \, dx. \]

Step 3: Use Additivity Property of Integrals

The integral from 0 to 4 can be split into two parts: \( \int_0^4 h(x) \, dx = \int_0^2 h(x) \, dx + \int_2^4 h(x) \, dx \). Thus, \( \int_2^4 h(x) \, dx = \int_0^4 h(x) \, dx - \int_0^2 h(x) \, dx \).

Step 4: Calculate \( \int_{2}^{4} f(x) \, dx \)

Use the additivity property: \( \int_{2}^{4} f(x) \, dx = \int_{0}^{4} f(x) \, dx - \int_{0}^{2} f(x) \, dx = 5 - (-3) = 8 \).

Step 5: Calculate \( \int_{2}^{4} g(x) \, dx \)

Similarly, for \( g(x) \), we have: \( \int_{2}^{4} g(x) \, dx = \int_{0}^{4} g(x) \, dx - \int_{0}^{2} g(x) \, dx = -1 - 2 = -3 \).

Step 6: Compute the Final Integral Using Step 2 Results

Substitute the results from Steps 4 and 5 back into the expression from Step 2: \[ \int_{2}^{4} (4f(x) - 3g(x)) \, dx = 4 \times 8 - 3 \times (-3) = 32 + 9 = 41. \]

Key Concepts

Properties of Integrals

Understanding the properties of integrals is essential when solving problems involving definite integrals. Definite integrals calculate the area under the curve of a function between two specific points, known as the limits of integration.
Some important properties to remember include:

• Linearity: This property allows you to simplify complex integrals by breaking them down into sums or differences of multiple, simpler integrals.
• Additivity: With this property, you can split the integral over a larger interval into the sum of two or more integrals over smaller intervals.
• Bounds: Changing the limits of integration can affect the result, and switching the limits changes the sign of the integral.

These properties powerfully simplify the integration process, making it easier to tackle difficult problems.

Linearity of Integrals

The linearity of integrals is a significant tool for handling multiple functions within a single integral. When dealing with an integral of the form \[ \int (c_1 f(x) + c_2 g(x)) \, dx, \] linearity tells us that it can be split into separate integrals as follows:\[ \int c_1 f(x) \, dx + \int c_2 g(x) \, dx. \]
What this means is that you can factor constants out of the integral and manage each function separately:

• Separate Functions: Linearity allows you to treat each function individually, simplifying calculations.
• Constant Multiplication: Multiplicative constants are pulled out of the integral, simplifying the integration of each function.

By understanding and applying linearity, you can simplify and solve integrals involving multiple functions more efficiently.

Additivity of Integrals

Additivity is a useful property that allows the division of an integral over an interval into the sum of integrals over sub-intervals. For a function \( h(x) \), the integral over \([a, c]\) can be split into:\[ \int_a^c h(x) \, dx = \int_a^b h(x) \, dx + \int_b^c h(x) \, dx. \]
This property is particularly helpful when:

• Combining Intervals: Additivity allows local segments to be analyzed separately before linking them back together.
• Simplifying Calculations: Long intervals are broken into simpler calculations, often making it easier to deal with known integrals.

Exploring the additivity of integrals can significantly streamline calculations by segmenting complex intervals into manageable pieces.

Calculating Integrals

Calculating integrals involves applying mathematical techniques and properties to determine the area under a curve. Breaking a complex expression like \[ \int (4f(x) - 3g(x)) \, dx \] into simpler parts using linearity makes it easier to compute.
Follow these steps:

• Apply Linearity: Separate the integral into components \( 4\int f(x) \, dx \) and \( -3\int g(x) \, dx \); this streamlines the process.


• Use Additivity: For each part, employ additivity to break down the interval if necessary \([0,4]\) into \([0,2]\) and \([2,4]\). This helps in using known values or simplifying functions.


• Substitute Known Values: Insert given or previously calculated values of integrals into your separated expressions.


• Compute: Combine results into a final value by evaluating and adding results as shown.

By systematically applying these steps, you can confidently calculate the integral's value, even for complex integrals.