Indefinite Integrals: Videos & Practice Problems

In calculus, the concept of the indefinite integral is closely related to antiderivatives, serving as the reverse process of differentiation. The notation for an indefinite integral is represented by an elongated 'S' symbol, indicating the operation of finding the antiderivative of a function, known as the integrand. When integrating a function \( f(x) \), the result is the antiderivative \( F(x) \) plus a constant \( C \), referred to as the constant of integration. This constant arises because the derivative of any constant is zero, meaning multiple functions can yield the same derivative.

The variable of integration, denoted as \( dx \), specifies the variable with respect to which the integration is performed. For example, if integrating with respect to \( x \), the integrand might be expressed as \( 3x^2 \) or \( 2x \). The process of finding the indefinite integral mirrors that of finding the antiderivative, with the addition of this new notation.

To illustrate the concept, consider the following examples:

1. **Indefinite Integral of 0**: The integral of 0 is simply a constant, represented as \( C \). This is because the derivative of any constant is zero, confirming that the antiderivative of 0 is indeed a constant.

2. **Indefinite Integral of a Constant**: For the integral of a constant, such as 3, the result is obtained by multiplying the constant by the variable of integration. Thus, the indefinite integral of 3 with respect to \( x \) is \( 3x + C \). This follows from the principle that the derivative of \( 3x \) is 3.

3. **Indefinite Integral of a Polynomial**: When integrating a polynomial like \( 3x^2 \), one can apply the power rule in reverse. The antiderivative of \( 3x^2 \) is \( x^3 + C \), since the derivative of \( x^3 \) yields \( 3x^2 \).

To verify the correctness of these integrals, one can differentiate the resulting functions. For instance, differentiating \( C \) gives 0, confirming the first example. The derivative of \( 3x + C \) is 3, matching the second example, and the derivative of \( x^3 + C \) is \( 3x^2 \), consistent with the third example. This process of checking ensures that the indefinite integrals have been computed accurately.

Understanding these principles of indefinite integrals is essential for further studies in calculus, as they form the foundation for more complex integration techniques and applications.

Indefinite integrals are a fundamental concept in calculus, representing the reverse process of differentiation. Understanding how to compute these integrals can be simplified by applying the power rule, which is analogous to the power rule used for derivatives. The power rule for derivatives states that for a function \( f(x) = x^n \), the derivative is given by:

\[ f'(x) = n \cdot x^{n-1} \]

In contrast, the power rule for integrals involves increasing the exponent by one and then dividing by the new exponent. Specifically, for the integral of \( x^n \), the formula is:

\[ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \]

Here, \( C \) represents the constant of integration, which is essential when calculating indefinite integrals. It is important to note that this rule is valid only when \( n \neq -1 \), as this would lead to division by zero.

To illustrate the application of the power rule for integrals, consider the integral of \( x^6 \). Using the power rule, we increase the exponent:

\[ \int x^6 \, dx = \frac{x^{6+1}}{6+1} + C = \frac{x^7}{7} + C \]

Similarly, for the function \( g(t) = t^4 \), the integral can be computed as follows:

\[ \int t^4 \, dt = \frac{t^{4+1}}{4+1} + C = \frac{t^5}{5} + C \]

These examples demonstrate the straightforward nature of applying the power rule for integration. To verify the correctness of the integral, one can differentiate the result. For instance, differentiating \( \frac{t^5}{5} + C \) will yield the original function \( t^4 \), confirming that the integration process was executed correctly.

In summary, mastering the power rule for integrals allows for efficient computation of indefinite integrals involving polynomial functions, reinforcing the relationship between differentiation and integration in calculus.

In the study of indefinite integrals, understanding how to integrate various functions is crucial. When faced with a combination of functions, such as polynomials, it’s essential to apply specific rules that simplify the integration process. The sum and difference rules, which you may recall from derivatives, are equally applicable to integrals. This means that if you have two functions being added or subtracted, you can integrate them separately. For example, the integral of \(x + 6\) can be expressed as the sum of the integrals of \(x\) and \(6\):

\[\int (x + 6) \, dx = \int x \, dx + \int 6 \, dx\]

Using the power rule for integration, the integral of \(x\) becomes \(\frac{x^2}{2}\), and the integral of a constant \(6\) results in \(6x\). Therefore, the complete integral is:

\[\int (x + 6) \, dx = \frac{x^2}{2} + 6x + C\]

Here, \(C\) represents the constant of integration, which is included because the integral of a function can yield an infinite number of antiderivatives differing by a constant. It’s important to note that you only need one \(C\) at the end, as any constants from separate integrals can be combined into a single constant.

Another important rule is the constant multiple rule, which states that if a constant multiplies a function, you can factor the constant out of the integral. For instance, when integrating \(5x^3\), you can rewrite it as:

\[\int 5x^3 \, dx = 5 \int x^3 \, dx\]

Applying the power rule again, the integral of \(x^3\) is \(\frac{x^4}{4}\), leading to:

\[\int 5x^3 \, dx = 5 \cdot \frac{x^4}{4} + C = \frac{5x^4}{4} + C\]

To integrate more complex functions, such as \(x^2 - 3x + 6\), you can apply both the sum and difference rule and the constant multiple rule. Start by breaking it down into separate integrals:

\[\int (x^2 - 3x + 6) \, dx = \int x^2 \, dx - \int 3x \, dx + \int 6 \, dx\]

Next, apply the constant multiple rule to the term \(-3x\) and integrate each term individually:

\[= \frac{x^3}{3} - 3 \cdot \frac{x^2}{2} + 6x + C\]

Combining these results gives:

\[\int (x^2 - 3x + 6) \, dx = \frac{x^3}{3} - \frac{3x^2}{2} + 6x + C\]

This process illustrates how to effectively use the sum and difference rule along with the constant multiple rule to tackle more complicated integrals. By breaking down the functions and applying these rules, you can simplify the integration of polynomials and other functions with ease.

To find the total profit function \( p(x) \) from the marginal profit function \( p'(x) \), we start by recognizing that \( p'(x) \) represents the derivative of the profit function. The process to obtain \( p(x) \) involves integrating \( p'(x) \).

Given the marginal profit function \( p'(x) = 0.03x - 1.5 \), we can express the total profit function as:

\[p(x) = \int (0.03x - 1.5) \, dx\]

We can split the integral into two parts:

\[p(x) = \int 0.03x \, dx - \int 1.5 \, dx\]

Next, we can factor out the constant from the first integral:

\[p(x) = 0.03 \int x \, dx - 1.5 \int 1 \, dx\]

Using the power rule for integration, where \(\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\), we find:

\[\int x \, dx = \frac{x^2}{2}\]

Thus, we can substitute this back into our equation:

\[p(x) = 0.03 \left(\frac{x^2}{2}\right) - 1.5x + C\]

Now simplifying the first term gives:

\[p(x) = 0.0015x^2 - 1.5x + C\]

To determine the constant \( C \), we use the condition \( p(0) = 0 \). Plugging in \( x = 0 \) into the profit function yields:

\[p(0) = 0.0015(0)^2 - 1.5(0) + C = C\]

Since \( p(0) = 0 \), it follows that \( C = 0 \). Therefore, the total profit function simplifies to:

\[p(x) = 0.0015x^2 - 1.5x\]

This function represents the total profit from selling parachutes, where \( x \) is the number of units sold. The quadratic term indicates that profit changes with the number of units sold, while the linear term reflects the fixed cost per unit.

The marginal revenue function for a ceramics company is represented as \( r'(x) \), which indicates the revenue generated per unit sold based on the quantity \( x \). To find the total revenue function \( r(x) \), we need to integrate the marginal revenue function. Given that \( r(0) = 0 \), we can determine the constant of integration later.

To start, we express the marginal revenue function as follows:

\[r'(x) = 1.68x^2 + x + 2.5\]

Next, we integrate each term separately:

\[r(x) = \int (1.68x^2 + x + 2.5) \, dx\]

Applying the rules of integration, we can break this down:

\[r(x) = \int 1.68x^2 \, dx + \int x \, dx + \int 2.5 \, dx\]

We can factor out the constant \( 1.68 \) from the first integral:

\[r(x) = 1.68 \int x^2 \, dx + \int x \, dx + 2.5 \int 1 \, dx\]

Now, applying the power rule for integration, where \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \), we find:

\[\int x^2 \, dx = \frac{x^{3}}{3}, \quad \int x \, dx = \frac{x^{2}}{2}, \quad \text{and} \quad \int 1 \, dx = x\]

Substituting these results back into the equation gives:

\[r(x) = 1.68 \cdot \frac{x^3}{3} + \frac{x^2}{2} + 2.5x + C\]

Calculating \( 1.68 \div 3 \) results in \( 0.56 \), so we can rewrite the revenue function as:

\[r(x) = 0.56x^3 + 0.5x^2 + 2.5x + C\]

To find the constant \( C \), we use the condition \( r(0) = 0 \). Substituting \( x = 0 \) into the revenue function results in:

\[r(0) = 0.56(0)^3 + 0.5(0)^2 + 2.5(0) + C = C\]

Since \( r(0) = 0 \), it follows that \( C = 0 \). Therefore, the total revenue function simplifies to:

\[r(x) = 0.56x^3 + 0.5x^2 + 2.5x\]

This function represents the total revenue generated by selling \( x \) units of ceramics, capturing the relationship between quantity sold and revenue earned.