Problem 55 - Calculus for Biology and Medicine

Question

Compute the indefinite integrals. $$ \int(x-2)(3-x) d x $$

Step-by-Step Solution

Answer

\(-\frac{x^3}{3} + \frac{5x^2}{2} - 6x + C\)

Step 1: Expand the Expression

Multiply the terms inside the parenthesis. We have \((x - 2)(3 - x) = x(3 - x) - 2(3 - x)\). Expanding this, we get:\[x(3 - x) = 3x - x^2\]and\[-2(3 - x) = -6 + 2x\].Combining these, we get:\[3x - x^2 - 6 + 2x = -x^2 + 5x - 6\].

Step 2: Integrate Term by Term

Integrate each term of the polynomial separately. Recall the power rule for integration: \[\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\]for any real number \(n eq -1\).1. Integrate \(-x^2\):\[ \int -x^2 \, dx = -\frac{x^3}{3} \]2. Integrate \(5x\):\[ \int 5x \, dx = \frac{5x^2}{2} \]3. Integrate \(-6\):\[ \int -6 \, dx = -6x \]

Step 3: Combine the Integrated Terms

Combine the results from each term's integration:\[ -\frac{x^3}{3} + \frac{5x^2}{2} - 6x + C\]where \(C\) is the constant of integration.

Key Concepts

Power Rule in Integration

The power rule is a fundamental principle in calculus that helps us integrate polynomials easily. When dealing with indefinite integrals of the form \( \int x^n \, dx \), the power rule allows us to find the antiderivative quickly. To use it, increment the exponent by one and divide by the new exponent:

• For example, for \( n = 2 \), we have \( \int x^2 \, dx = \frac{x^{2+1}}{2+1} = \frac{x^3}{3} \).
• If the term includes a constant, like \( 5x^2 \), apply the rule to the \( x^2 \) and retain the constant: \( \int 5x^2 \, dx = 5 \cdot \frac{x^3}{3} = \frac{5x^3}{3} \).
• Always remember, this rule only works for \( n eq -1 \). The case with \( n = -1 \) is special and leads to a logarithmic function.

By utilizing the power rule, we simplify the process of finding indefinite integrals, making it a handy tool for tackling integration problems without hassle.

Understanding Integration

Integration is a core concept in calculus, essential for finding areas under curves and solving differential equations. Specifically, it's the reverse process of differentiation. When we integrate, we're essentially looking for a function whose derivative matches the given function. In indefinite integration, like the one present in our problem, the goal is to determine the general form of an antiderivative. This involves finding a function that describes a whole family of curves. To distinguish these curves, we add a constant \( C \), representing any vertical shift. This step is why the constant of integration is immediately visible in the solution.Consider these steps during integration:

• Identify individual terms in the polynomial to apply rules separately.
• Use known integration formulas, like the power rule for polynomials.
• Always include the constant \( C \) once integration is completed to represent all possible antiderivatives.

Thus, integration can be viewed as piecing together parts of a puzzle to find the broader picture of the function you're working with.

Polynomial Expansion

Polynomial expansion involves transforming a polynomial expressed in a product form into a sum form. It is a crucial step in simplifying expressions ahead of integration. Before you can apply integration rules to a problem like \( \int(x-2)(3-x) \), it’s important to expand the expression fully.Here's a rundown of the expansion process:

• First, use distributive property: multiply each term in the first parenthesis by every term in the second. For \( (x-2)(3-x) \), do \( x \cdot (3-x) \) and \(-2 \cdot (3-x) \).
• Simplify each multiplication: \( x \cdot 3 - x \cdot x = 3x - x^2 \) and \(-2 \cdot 3 + 2 \cdot x = -6 + 2x \).
• Add the simplified results together: \( 3x - x^2 - 6 + 2x \) simplifies further to \(-x^2 + 5x - 6 \).

Polynomial expansion allows you to then apply integration rules to each term effectively. Remember, correctly expanding polynomials is a gateway to simplifying further calculus operations such as finding derivatives or integrals.