Problem 56 - Calculus for Biology and Medicine

Question

Compute the indefinite integrals. $$ \int(x+1) x^{2} d x $$

Step-by-Step Solution

Answer

The integral is \(\frac{x^4}{4} + \frac{x^3}{3} + C\).

Step 1: Expand the Expression

First, expand the expression \((x+1)x^2\). Use the distributive property to do this: \((x+1)x^2 = x \cdot x^2 + 1 \cdot x^2 = x^3 + x^2\). This results in a new integral \(\int (x^3 + x^2) \, dx\).

Step 2: Separate the Integral

Separate the integrals of each term within the expression. This gives us: \(\int x^3 \, dx + \int x^2 \, dx\). This will allow us to integrate each term individually.

Step 3: Integrate Each Term

Now, integrate each term separately. For \( \int x^3 \, dx\), use the power rule of integration: \(\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\). This results in \( \frac{x^4}{4} \). For \( \int x^2 \, dx\), similarly apply the power rule to get: \(\frac{x^3}{3} \).

Step 4: Combine the Integrated Terms

Combine the results of the integral into a single expression: \(\frac{x^4}{4} + \frac{x^3}{3} + C\), where \(C\) is the constant of integration.

Key Concepts

Power Rule

The power rule is a straightforward method used to find the antiderivative or indefinite integral of a function involving a power of a variable. It is a fundamental tool in calculus. The power rule states: \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \), where \( n eq -1 \) and \( C \) is the constant of integration.

This rule is highly useful when dealing with polynomials, as each term can be integrated individually. For example, if you have \( \int x^3 \, dx \), by applying the power rule, we find the antiderivative is \( \frac{x^4}{4} + C \).

• It is important to increase the exponent by one.
• Divide by the new exponent.
• Add the constant of integration \( C \).

Power rule simplifies the process of integration, making it a fundamental building block for more complex problems.

Integration by Parts

Integration by parts is a powerful technique used when integrating the product of two functions, which isn't straightforward using standard methods. Although not directly used in the given exercise, it is beneficial to understand for more complex problems.

The technique is derived from the product rule for differentiation and is expressed as:

\[ \int u \, dv = uv - \int v \, du \]

Here, you typically choose \( u \) to be a function that simplifies when differentiated, and \( dv \) as a function that simplifies when integrated. Applying integration by parts can involve multiple steps and requires careful selection of \( u \) and \( dv \).

• Choose \( u \) and \( dv \) wisely.
• Differentiate \( u \) to find \( du \), integrate \( dv \) to find \( v \).
• Substitute into the formula and simplify.

Though not needed in this particular problem, mastering integration by parts allows tackling a wide array of integrals, enhancing your problem-solving arsenal.

Distributive Property

The distributive property is a key algebraic rule used for multiplying a single term across a sum or difference of terms. It is particularly useful in preparing expressions for integration by separating and simplifying them.

In the original problem, \( (x+1)x^2 \) is expressed using the distributive property as \( x \cdot x^2 + 1 \cdot x^2 = x^3 + x^2 \). This step is crucial because it breaks down the polynomial, allowing the application of the power rule to each term individually.

The distributive property can be stated as:

• If \( a(b + c) = ab + ac \)
• It ensures each term in the parentheses is multiplied by the term outside.

This strategy is helpful when preparing integrals, and its utility extends beyond pure algebraic simplification to serve the needs of calculus, helping simplify functions into integrable pieces.