Problem 44 - Calculus for Biology and Medicine

Question

Compute the indefinite integrals. $$ \int\left(\frac{1}{2} x^{5}+2 x^{3}-1\right) d x $$

Step-by-Step Solution

Answer

\( \frac{x^6}{12} + \frac{x^4}{2} - x + C \)

Step 1: Identify the Components of the Integral

The given integral is \( \int\left(\frac{1}{2} x^{5}+2 x^{3}-1\right) d x \). We see that there are three terms inside the integral: \( \frac{1}{2} x^5 \), \( 2x^3 \), and \( -1 \). We will integrate each term separately.

Step 2: Integrate the First Term

The first term is \( \frac{1}{2} x^5 \). To integrate this term, apply the power rule for integration, which states that \( \int x^n dx = \frac{x^{n+1}}{n+1} + C \). Thus, \( \int \frac{1}{2} x^5 dx = \frac{1}{2} \cdot \frac{x^{6}}{6} = \frac{x^6}{12} \).

Step 3: Integrate the Second Term

The second term is \( 2x^3 \). Using the power rule, we have \( \int 2x^3 dx = 2 \cdot \frac{x^{4}}{4} = \frac{x^4}{2} \).

Step 4: Integrate the Third Term

The third term is \( -1 \). The integral of a constant \( c \) is \( cx + C \) (where \( C \) is the constant of integration). Therefore, \( \int -1 dx = -x \).

Step 5: Combine the Integrated Terms

Combine the results of the integrated terms from Steps 2, 3, and 4. This gives the indefinite integral: \( \int\left(\frac{1}{2} x^{5}+2 x^{3}-1\right) dx = \frac{x^6}{12} + \frac{x^4}{2} - x + C \), where \( C \) is the constant of integration.

Key Concepts

Understanding the Power Rule

The power rule is a fundamental technique used in calculus for performing integration. It allows us to integrate terms of the form \( x^n \). This is a crucial step in finding indefinite integrals. The power rule for integration states:

• \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \)

This formula involves adding 1 to the exponent \( n \), then dividing by the new exponent \( n + 1 \). Remember, the power rule only applies when \( n eq -1 \), as dividing by zero is undefined.
Any constant multiplier of \( x^n \) should be carried along during integration. For instance, with the term \( \frac{1}{2} x^5 \), we treat \( \frac{1}{2} \) as a constant. Thus, the integration process looks like:

• First, apply the power rule to \( x^5 \).
• Then multiply by the constant, \( \frac{1}{2} \).
• The result is \( \frac{x^6}{12} \).

Using the power rule makes integration systematic and straightforward. It’s like a blueprint for integrating polynomial terms.

The Process of Integration

Integration is basically the reverse process of differentiation. It's a method to find areas under curves or to recover a function from its derivative. The indefinite integral represents a family of functions and is expressed as the integral symbol followed by a function and \( dx \). Consider the given integral:

• \( \int\left( \frac{1}{2} x^5 + 2 x^3 - 1 \right) \, dx \)

This tells us to find the antiderivative of the polynomial given. To execute this, we:

• Break down the integral into components that are easier to work with.
• Apply the power rule to each term separately.
• Combine the antiderivatives to get the overall integral.

By integrating each term:

• First term: \( \frac{x^6}{12} \)
• Second term: \( \frac{x^4}{2} \)
• Third term: \( -x \)

When combined, these lead us to the complete antiderivative plus a constant of integration:

• \( \int\left( \frac{1}{2} x^5 + 2 x^3 - 1 \right) \, dx = \frac{x^6}{12} + \frac{x^4}{2} - x + C \)

Integration simplifies complex problems by unraveling them into simpler components.

Role of the Constant of Integration

In calculus, when we find an indefinite integral, we add a constant \( C \) to represent all possible constant values that could be part of the antiderivative. This is crucial because integrating a function's derivative only gives back part of the original function. The constant of integration acknowledges that multiple functions could have the same derivative. Here’s why it matters:

• If \( f'(x) = g'(x) \), then \( f(x) \) and \( g(x) \) differ by a constant.
• The general solution becomes \( F(x) + C \).
• Neglecting \( C \) can lead to incomplete or incorrect solutions.

For instance, when we integrated the given polynomial:

• Every term is integrated separately.
• The results are combined along with \( C \) to cover all potential original forms.
• The final expression is \( \frac{x^6}{12} + \frac{x^4}{2} - x + C \).

Thus, including \( C \) ensures our solution represents every potential antiderivative. It plays a pivotal role in capturing the entire family of solutions for an indefinite integral.