Problem 41 - Applied Calculus

Question

Find the indefinite integrals. $$ \int\left(t^{3}+6 t^{2}\right) d t $$

Step-by-Step Solution

Answer

\( \int (t^3 + 6t^2) \, dt = \frac{t^4}{4} + 2t^3 + C \)

Step 1: Separate the Integral

Start by breaking down the integral into separate integrals for each term within the function: \( \int (t^3 + 6t^2) \, dt = \int t^3 \, dt + \int 6t^2 \, dt \).

Step 2: Integrate Each Term

Use the power rule for integration, \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \), adding a constant \( C \) for indefinite integrals. For the first integral, apply the rule to \( t^3 \): \( \int t^3 \, dt = \frac{t^{3+1}}{3+1} + C_1 = \frac{t^4}{4} + C_1 \). For the second integral, apply the rule to \( 6t^2 \): \( \int 6t^2 \, dt = 6 \cdot \frac{t^{2+1}}{2+1} + C_2 = 2t^3 + C_2 \).

Step 3: Combine the Results

Write the combined expression of the results from step 2, incorporating the constants into a single constant functionality \( C \): \( \int (t^3 + 6t^2) \, dt = \frac{t^4}{4} + 2t^3 + C \).

Key Concepts

Power Rule for Integration

When it comes to integrating functions where each term is a power of a variable, the power rule for integration is incredibly helpful. This rule allows us to handle each term separately and integrate it in a straightforward manner.

The power rule states that the integral of a power function \( x^n \) is \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \), where \( C \) is the constant of integration. This applies as long as \( n eq -1 \), since dividing by zero in calculus is undefined.

For example, in the exercise, you apply the power rule individually to \( t^3 \) and \( 6t^2 \). For \( t^3 \), you get \( \frac{t^{3+1}}{3+1} = \frac{t^4}{4} \). Similarly, for \( 6t^2 \), the result is \( 6 \times \frac{t^{2+1}}{2+1} = 2t^3 \). By applying this rule, you're able to effectively integrate each term with ease.

Constant of Integration

In the world of indefinite integrals, the constant of integration, denoted as \( C \), holds significant importance. Whenever you find the indefinite integral of a function, unless additional conditions are specified, you must always remember to add \( C \).

This constant represents the family of functions that are solutions to the integral. Every differentiable function related to an antiderivative can differ by a constant. Therefore, without further information like initial conditions or boundaries, the integration process leaves us with this constant.

Live illustration: When integrating \( t^3 \) to obtain \( \frac{t^4}{4} \), you add a constant \( C_1 \). Similarly, when integrating \( 6t^2 \) to get \( 2t^3 \), you add \( C_2 \). Combining these constants just gives us a general \( C \) when writing the complete solution, \( \frac{t^4}{4} + 2t^3 + C \). This traditionally simplifies the expression to universally signify all potential solutions.

Definite vs. Indefinite Integrals

Fundamentally, integrals can be categorized into two primary types: definite and indefinite integrals. Understanding the difference is crucial in grasping how they are applied in mathematics.

Indefinite integrals, like the one given in the exercise \( \int (t^3 + 6t^2) \, dt \), aim to find the general form of the antiderivative. These don't have specific limits or bounds, which is why the constant of integration \( C \) is added, due to the absence of further conditions limiting the solutions.

On the other hand, definite integrals come with specified upper and lower limits. For instance, \( \int_{a}^{b} f(t) \, dt \). Computing a definite integral will give you a specific numerical value, which represents the area under the curve of the function between the limits \( a \) and \( b \). Here, there is no need to add \( C \), as the integral evaluates to a precise number rather than a general function expression.

Knowing when to employ indefinite versus definite integrals can significantly affect the outcomes of your integration and the breadth of insights you can derive from them.