Problem 53 - Calculus for Biology and Medicine

Question

Compute the indefinite integrals. $$ \int(x-1)(x+1) d x $$

Step-by-Step Solution

Answer

The indefinite integral is \(\frac{x^3}{3} - x + C\).

Step 1: Expand the integrand

To simplify the integration, start by expanding the product in the integrand. Multiply \(x - 1\) and \(x + 1\): \[(x - 1)(x + 1) = x^2 + x - x - 1^2 = x^2 - 1\]So, \int(x-1)(x+1) dx = \int(x^2 - 1) dx.\.

Step 2: Integrate each term separately

Integrate each term of \(x^2 - 1\) separately:\[\int (x^2 - 1) dx = \int x^2 dx - \int 1 dx.\]Each term is integrated individually.

Step 3: Integrate \\int x^2 dx\\

Use the power rule for integration, which states \int x^n dx = \frac{x^{n+1}}{n+1} + C\, where \C\ is a constant of integration. Applying to \(x^2\):\[\int x^2 dx = \frac{x^{3}}{3} + C_1.\]

Step 4: Integrate \\int 1 dx\\

The integral of a constant \1\ is simply \x\, as each unit of \x\ provides 1:\[\int 1 dx = x + C_2.\]

Step 5: Combine the results

Combine the integrated results from Steps 3 and 4:\[\int(x^2 - 1) dx = \frac{x^3}{3} - x + C.\]Here, \C\ is the combined constant of integration from \C_1\ and \C_2\.

Key Concepts

Expanding Integrands

When faced with integrals, particularly those involving products of expressions, it's often beneficial to simplify the expression before integration. This process is known as 'expanding the integrands.' In our original exercise, we are given the expression \((x-1)(x+1)\) inside the integral.
By expanding this product, we transform it into a simpler form that is easier to integrate. Applying the distributive property to \((x-1)(x+1)\) yields:

• Multiply the first terms: \(x \cdot x = x^2\)
• Multiply the outer terms: \(x \cdot 1 = x\)
• Multiply the inner terms: \(-1 \cdot x = -x\)
• Multiply the last terms: \(-1 \cdot 1 = -1\)

Combine these results: \[x^2 + x - x - 1 = x^2 - 1\] Now, the integral becomes \(\int (x^2 - 1) \, dx\), a far simpler expression to work with. This simplification paves the way for efficient and straightforward application of integration rules.

Power Rule for Integration

Integrating polynomials is a foundational skill in calculus, often involving the 'power rule for integration'. This rule lets us integrate terms of the form \(x^n\). It is particularly handy when we encounter expanded expressions like \(x^2 - 1\). The power rule states: \[ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \] where \(n\) is a real number not equal to -1, and \(C\) represents the constant of integration.
For our exercise, using the power rule on \(x^2\) involves raising the power by one and dividing by the new power:

• \(\int x^2 \, dx = \frac{x^{3}}{3} + C_1\)

Each term in the polynomial is treated this way, allowing us to separately integrate each term effectively before combining them. This rule simplifies the integration process substantially, converting each term into an easily integrated fraction.

Constant of Integration

In indefinite integrals, you will often see a term denoted by \(C\), known as the 'constant of integration'. This concept captures the idea that an integral represents a family of functions, each differing by a constant.
When integrating, every constant function's derivative is zero, thus not affecting the underlying equation. Therefore, when you find an antiderivative, there are infinitely many forms of it, each differing from the others by a constant.
For our problem, after integrating using the power rule, we obtained

• \(\int x^2 \, dx = \frac{x^3}{3} + C_1\)
• \(\int 1 \, dx = x + C_2\)

Upon combining these results to \(\int (x^2 - 1) \, dx = \frac{x^3}{3} - x + C\), the combined \(C\) encompasses both \(C_1\) and \(C_2\). This makes sure that our solution incorporates all possible vertical shifts of the function graph. Remember that including the constant of integration is essential to fully represent the solution space of antiderivatives in calculus.