Problem 98 - Calculus Volume 1

Question

In the following exercises, given that \(\int_{0}^{1} x d x=\frac{1}{2}, \int_{0}^{1} x^{2} d x=\frac{1}{3}, \quad\) and \(\quad \int_{0}^{1} x^{3} d x=\frac{1}{4}\) compute the integrals. \(\int_{0}^{1}\left(1+x+x^{2}+x^{3}\right) d x\)

Step-by-Step Solution

Answer

The integral is \(\frac{25}{12}\).

Step 1: Understand the Integral

We need to compute the integral of the function \(1+x+x^{2}+x^{3}\) from 0 to 1, which is written as: \(\int_{0}^{1}(1+x+x^{2}+x^{3}) \, dx\). The goal is to integrate the function over the interval [0,1].

Step 2: Split the Integral

Split the given integral into the sum of individual integrals. We can rewrite it as: \(\int_{0}^{1} 1 \, dx + \int_{0}^{1} x \, dx + \int_{0}^{1} x^{2} \, dx + \int_{0}^{1} x^{3} \, dx\).

Step 3: Calculate Each Integral

Calculate each integral separately:1. \(\int_{0}^{1} 1 \, dx = [x]_{0}^{1} = 1 - 0 = 1\)2. \(\int_{0}^{1} x \, dx = \frac{1}{2}\)3. \(\int_{0}^{1} x^{2} \, dx = \frac{1}{3}\)4. \(\int_{0}^{1} x^{3} \, dx = \frac{1}{4}\)

Step 4: Add the Resultant Integrals

Add the results from each integral to find the total. Therefore, the sum is:\(1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} = \) Now, find a common denominator and add them.Find the least common denominator (LCD) for 1, 2, 3, 4: The LCD is 12.Rewriting each term with denominator 12:\(1 = \frac{12}{12}\), \(\frac{1}{2} = \frac{6}{12}\), \(\frac{1}{3} = \frac{4}{12}\), \(\frac{1}{4} = \frac{3}{12}\).Now add them: \(\frac{12}{12} + \frac{6}{12} + \frac{4}{12} + \frac{3}{12} = \frac{25}{12}\).

Key Concepts

Integration Techniques

Integration is a fundamental concept in calculus, often used to find areas under curves and solve differential equations. A vital technique in integration is splitting a complex integral into simpler parts. This is known as the additive property of integrals.

To integrate the polynomial function \(1 + x + x^2 + x^3\) over the interval [0, 1], we can handle it piece by piece. This approach involves evaluating each term of the polynomial separately. You'll get:

• \( \int_{0}^{1} 1 \, dx \)
• \( \int_{0}^{1} x \, dx \)
• \( \int_{0}^{1} x^2 \, dx \)
• \( \int_{0}^{1} x^3 \, dx \)

By integrating each part and then summing the results, the problem becomes more manageable. The method illustrates the power of breaking down integrals, simplifying calculations and enhancing understanding of how integration works with polynomial structures.

Polynomial Functions

Polynomial functions, like \(1 + x + x^2 + x^3\), are expressions made up of variables raised to non-negative integer powers. These are incredibly common in calculus due to their smooth, continuous nature. To integrate polynomial functions, each term is considered separately based on the power of \(x\). For instance, when you address higher powers of \(x\), like \(x^2\) or \(x^3\), traditional power rule of integration comes into play. Given formula, the rule is:

• \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \)

For the interval [0,1], calculus problems often become a straightforward task because the terms like constants and linear \(x\) terms have established solutions such as constant integrals (for constant functions) which lead to interesting results with nice fractional outputs. Thus, polynomial functions present a clear, logical expression to work with, making them an essential building block in calculus.

Calculus Problem Solving

Calculus problem solving often involves combining multiple calculus concepts to address a specific problem. In the context of definite integrals, like the one given in the exercise, it involves understanding the function, decomposing it into simpler parts, calculating each part, and finally combining the solutions.
Solving \( \int_{0}^{1} (1 + x + x^2 + x^3) \, dx \) means:

• Recognizing the polynomial's structure and its components.
• Applying integration rules separately to each component.
• Utilizing arithmetic to add fractions with a common denominator.

At its core, calculus problem-solving is a logical process. It involves practice and familiarity with rules like the power rule and properties of integrals, bringing abstract mathematical concepts down to more concrete steps. This comprehensive analysis of polynomial behavior within definite integrals exemplifies the procedural and methodical nature of solving calculus problems effectively.