Problem 27 - Calculus: Early Transcendentals

Question

Evaluate the integral. \( \displaystyle \int^1_0 (u + 2) (u - 3) \,du \)

Step-by-Step Solution

Answer

The integral evaluates to \(-\frac{37}{6}\).

Step 1: Expand the Expression Inside the Integral

First, expand the expression \((u + 2)(u - 3)\) using the distributive property. Apply the distributive property: \((u + 2)(u - 3) = u(u - 3) + 2(u - 3) = u^2 - 3u + 2u - 6\). Combine like terms:\(u^2 - u - 6\).

Step 2: Set Up the Integral with the Expanded Expression

Now that we have expanded the expression, we set up the integral to evaluate. Replace the original integrand with the expanded integrand:\(\int_0^1 (u^2 - u - 6) \, du\).

Step 3: Integrate Term by Term

Integrate each term of \(\int (u^2 - u - 6) \, du\) independently:- The integral of \(u^2\) is \(\frac{u^3}{3}\).- The integral of \(-u\) is \(-\frac{u^2}{2}\).- The integral of \(-6\) is \(-6u\).Thus, the integral becomes:\(\left[\frac{u^3}{3} - \frac{u^2}{2} - 6u\right]_0^1\).

Step 4: Evaluate the Indefinite Integral at the Bounds

Evaluate the result from Step 3 at the upper and lower bounds of the integral. First, plug in the upper bound (1): \(\frac{1^3}{3} - \frac{1^2}{2} - 6(1) = \frac{1}{3} - \frac{1}{2} - 6\). Now, plug in the lower bound (0): \(\frac{0^3}{3} - \frac{0^2}{2} - 6(0) = 0\).

Step 5: Simplify to Find the Final Answer

Now, subtract the value at the lower bound from the value at the upper bound:\(\left(\frac{1}{3} - \frac{1}{2} - 6\right) - (0)\).To simplify:Convert \(\frac{1}{3}\) and \(\frac{1}{2}\) to a common denominator:\(\frac{1}{3} = \frac{2}{6}, \frac{1}{2} = \frac{3}{6}\).So, \(\frac{2}{6} - \frac{3}{6} = -\frac{1}{6}\).Now, simplify:\(-\frac{1}{6} - 6 = -\frac{1}{6} - \frac{36}{6} = -\frac{37}{6}\). Final solution: \(-\frac{37}{6}\).

Key Concepts

Distributive Property

The distributive property is an essential mathematical concept that allows us to simplify and manipulate expressions. It states that multiplying a single term by two or more terms within parentheses is equivalent to multiplying the single term by each term inside the parentheses, and then summing the results. In our example, we used the distributive property to expand the expression \[(u + 2)(u - 3) = u(u - 3) + 2(u - 3)\]. Here, each term inside the first set of parentheses multiplies each term inside the second set of parentheses. This step-by-step expansion leads us to \[u^2 - 3u + 2u - 6\]. Subsequently, we combine like terms to compact our expression into \[u^2 - u - 6\]. Understanding and applying the distributive property correctly is crucial when working with polynomials and preparing them for further operations, like integration.

Integral Evaluation

Integral evaluation refers to the process of calculating the definite or indefinite integral of a function. In this exercise, we focus on definite integral evaluation, meaning we're integrating between specific bounds: 0 and 1. Once the polynomial \[u^2 - u - 6\]is set up as our integrand, we must integrate each term independently. Let's break these down:

• The integral of \(u^2\) becomes \(\frac{u^3}{3}\).
• The integral of \(-u\) becomes \(-\frac{u^2}{2}\).
• The integral of \(-6\) becomes \(-6u\).

Thus, the expression for the integral from Step 3 is \[\left[\frac{u^3}{3} - \frac{u^2}{2} - 6u\right]_0^1\].After finding the indefinite integral, evaluate it at the bounds to find the definite value: substitute 1 (upper bound) and 0 (lower bound) into the integrated expression and subtract to obtain the final result. This process yields our solution of \(-\frac{37}{6}\).

Polynomial Expansion

Polynomial expansion is a technique that involves rewriting a product of expressions as a sum or polynomial. This approach is beneficial in calculus when preparing functions for integration or differentiation. When given the product \[(u + 2)(u - 3)\],we expand it to turn it into a polynomial form that is easier to integrate. The steps taken include distributing each term of the first polynomial over each term of the second, as shown:

• Multiply \(u\) by \(u - 3\) to get \(u^2 - 3u\).
• Multiply \(2\) by \(u - 3\) to get \(2u - 6\).

The expanded polynomial is \(u^2 - u - 6\) after combining like terms. Understanding polynomial expansion not only aids in solving calculus problems but also deepens comprehension of algebraic manipulation, which is a foundation for more advanced mathematics.