Problem 19 - Thomas Calculus

Question

In Exercises \(17-56,\) find the most general antiderivative or indefinite integral. You may need to try a solution and then adjust your guess. Check your answers by differentiation. $$\int\left(3 t^{2}+\frac{t}{2}\right) d t$$

Step-by-Step Solution

Answer

The most general antiderivative is \( t^3 + \frac{1}{4}t^2 + C \).

Step 1: Understand the Problem

We need to find the most general antiderivative or indefinite integral of the expression \( \int \left(3t^2 + \frac{t}{2}\right) dt \). This involves integrating each term separately.

Step 2: Integrate Each Term Separately

The integral of \( 3t^2 \) is calculated using the power rule, which gives \( \frac{3}{3}t^3 = t^3 \). For the term \( \frac{t}{2} \), apply the power rule to get \( \frac{1}{2} \cdot \frac{1}{2+1}t^{2+1} = \frac{1}{4}t^2 \). Integrate each term separately.

Step 3: Combine the Antiderivatives

Combine the results of the separate integrals: \( t^3 + \frac{1}{4}t^2 \). Add the constant of integration, \( C \), to get the most general antiderivative: \( t^3 + \frac{1}{4}t^2 + C \).

Step 4: Differentiate to Check the Solution

Differentiate the expression \( t^3 + \frac{1}{4}t^2 + C \). The derivative of \( t^3 \) is \( 3t^2 \). The derivative of \( \frac{1}{4}t^2 \) is \( \frac{1}{2}t \), and the derivative of \( C \) is 0. Thus, the derivative is \( 3t^2 + \frac{t}{2} \), which matches the original integrand.

Key Concepts

Antiderivative

Finding an antiderivative is like performing reverse differentiation. We seek a function whose derivative gives us back the original function. In indefinite integration, each term of the expression is integrated separately. This means that we identify a function that, when differentiated, results in the term within the integral sign. For the integral \(\int \left(3t^2 + \frac{t}{2}\right) dt\), we determine individual functions whose derivatives are \(3t^2\) and \(\frac{t}{2}\) respectively. This process is key because it allows us to determine the general formula for the integral, known as the most general antiderivative.

Integration by Parts

Integration by parts is a technique derived from the product rule of differentiation. However, in this specific problem, it is not necessary because we can straightforwardly apply simpler integration rules. Integration by parts is useful when we encounter a product of two different functions or when one function is easily differentiated and the other is easily integrated. Fortunately, for the integral \(\int \left(3t^2 + \frac{t}{2}\right) dt\), both terms can be individually integrated using the power rule, simplifying the overall process without the need for integration by parts.

Constant of Integration

When we find the indefinite integral of a function, we always add a constant of integration, denoted by \( C \). This is because there are infinitely many antiderivatives differing only by a constant. The general form of an indefinite integral involves adding this constant to account for all possible constants that could have been present before differentiation.

• The antiderivative of \(3t^2\) and \(\frac{t}{2}\) result in \(t^3\) and \(\frac{1}{4}t^2\) respectively.
• Thus, the general solution is \(t^3 + \frac{1}{4}t^2 + C\), ensuring that all possible solutions are included.

This addition is crucial and must always be included when computing an indefinite integral.

Power Rule for Integration

The power rule for integration is one of the simplest and most important tools in calculus for finding antiderivatives. It provides a way to integrate terms of the form \(x^n\). The rule states that the integral of \(x^n\) is \(\frac{x^{n+1}}{n+1}\), provided that \(n eq -1\).

• In application, the term \(3t^2\) becomes \(\frac{3}{3}t^3 = t^3\) when using the power rule.
• Similarly, \(\frac{t}{2}\) can be rewritten as \(\frac{1}{2}t^1\). Using the power rule, it becomes \(\frac{1}{4}t^2\).

This rule simplifies the process of integration considerably and is essential in efficiently solving many integrals.