Problem 1 - Thomas Calculus

Question

Evaluate the integrals in Exercises \(1-28\). $$\int_{0}^{2} x(x-3) d x$$

Step-by-Step Solution

Answer

The value of the integral is -\(\frac{10}{3}\).

Step 1: Expand the Expression Inside the Integral

The first step is to simplify the expression inside the integral. Given the integral \(\int_{0}^{2} x(x-3) \, dx\), expand the terms: \(x(x-3) = x^2 - 3x\). The expression inside the integral becomes \(x^2 - 3x\).

Step 2: Set Up the New Integral

Now, set up the integral with the expanded expression: \(\int_{0}^{2} (x^2 - 3x) \, dx\). This will be solved next.

Step 3: Integrate Each Term Separately

Integrate each term separately within the expanded expression. - The first term: \(\int x^2 \, dx\) gives \(\frac{x^3}{3}\).- The second term: \(\int -3x \, dx\) gives \(-\frac{3x^2}{2}\).

Step 4: Evaluate the Definite Integral

First, we combine the integrated terms: \( \left[ \frac{x^3}{3} - \frac{3x^2}{2} \right]_{0}^{2} \).Now evaluate at the bounds:- Evaluate at 2: \( \left( \frac{2^3}{3} - \frac{3(2)^2}{2} \right) = \left( \frac{8}{3} - \frac{12}{2} \right) = \left( \frac{8}{3} - 6 \right) = \left( \frac{8}{3} - \frac{18}{3} \right) = -\frac{10}{3} \).- Evaluate at 0: \( \left( \frac{0^3}{3} - \frac{3(0)^2}{2} \right) = 0 \).Subtract: \(-\frac{10}{3} - 0 = -\frac{10}{3}\).

Key Concepts

Polynomial Integration

Polynomial integration is one of the essential techniques in calculus used to find the antiderivative of polynomial expressions. When we talk about polynomial integration, we are referring to the process of integrating expressions like \( ax^n \), where \( a \) is a constant and \( n \) is a whole number. To find the integral of a polynomial, you apply the power rule of integration:

• For \( x^n \), the integral is \( \frac{x^{n+1}}{n+1} + C \), where \( C \) is the constant of integration for indefinite integrals.

The power rule allows each term in the polynomial to be integrated separately, because each term is essentially a monomial. This simplifies the process considerably and is especially effective for straightforward polynomials like the one in our example: \( x(x-3) = x^2 - 3x \). By breaking it down to individual terms, polynomial integration becomes a task of applying this simple rule repeatedly.

Integral Bounds Evaluation

Evaluating definite integrals such as \( \int_{a}^{b} f(x) \, dx \) involves calculating the area under the curve represented by the function \( f(x) \) from \( x = a \) to \( x = b \). This process requires two main steps:1. **Integrate the Function**: First, you find the antiderivative, or the integral, of the function you wish to evaluate. This provides a new function that represents the area under the original curve. 2. **Apply the Bounds**: With the antiderivative \( F(x) \), you evaluate it at the upper and lower bounds, then subtract: \( F(b) - F(a) \).In the example \( \int_{0}^{2} x(x-3) \, dx \), once you have integrated the function to \( \frac{x^3}{3} - \frac{3x^2}{2} \), you simply plug in the upper limit \( x = 2 \) and lower limit \( x = 0 \). The difference \( F(2) - F(0) \) gives the area under the curve from 0 to 2. This process is a powerful tool in calculus, providing not just values, but insights into how areas and accumulations operate under curves.

Integration Techniques

Integration techniques refer to various methods used to solve integrals, each with its unique application depending on the function form. Simple polynomial expressions can be addressed with straightforward techniques like the power rule, as we've seen.For more complex functions, you might need:

• **Substitution Method**: Useful when an integral includes a function and its derivative. This method substitutes parts of the integral to simplify it.
• **Integration by Parts**: Valuable for products of functions and is based on the product rule of differentiation.
• **Partial Fractions**: Employed when dealing with rational functions, breaking them into simpler parts.

The example \( \int_{0}^{2} x(x-3) \, dx \) primarily uses direct polynomial techniques. Recognizing patterns or knowing when to apply specific integration strategies can greatly simplify solving integrals, making this area of calculus versatile and powerful for analyzing continuous functions.