Problem 59 - 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: Identify Integral Components

The function to integrate is \(t^3 + 6t^2\). Notice that it is a sum of two terms: \(t^3\) and \(6t^2\). Since integration respects the linearity property, we can integrate each term individually.

Step 2: Integrate the First Term

The first term is \(t^3\). To integrate \(t^n\), where \(n\) is any real number, use the formula: \(\int t^n \, dt = \frac{t^{n+1}}{n+1} + C\). Applying this, we have \(\int t^3 \, dt = \frac{t^{3+1}}{3+1} = \frac{t^4}{4}\).

Step 3: Integrate the Second Term

The second term is \(6t^2\). Treat the constant \(6\) as a coefficient that can be factored out. Hence, \(\int 6t^2 \, dt = 6 \int t^2 \, dt\). Use the integration formula: \(\int t^2 \, dt = \frac{t^{2+1}}{2+1} = \frac{t^3}{3}\). So, \(6 \times \frac{t^3}{3} = 2t^3\).

Step 4: Combine Results

Now, combine the integrated results from Step 2 and Step 3. The integral becomes \(\frac{t^4}{4} + 2t^3 + C\), where \(C\) is the constant of integration.

Key Concepts

Integration Techniques

Understanding integration techniques is key to solving indefinite integrals like the one given in the exercise. A fundamental technique is recognizing the linearity of integration, which allows us to split the integral of a sum into the sum of integrals. This means that for a function of the form \(f(t) = a(t^n) + b(t^m)\), we can write:

• \(\int f(t) \, dt = \int a(t^n) \, dt + \int b(t^m) \, dt\)

This property simplifies the integration process by allowing each term to be integrated separately.
Another technique involves using a basic power rule for integration. This rule states that for any term of the form \(t^n\), the integral is given by:

• \(\int t^n \, dt = \frac{t^{n+1}}{n+1} + C\)

where \(C\) is the constant of integration. Recognizing and applying these foundational techniques makes solving indefinite integrals more straightforward.

Calculus Fundamentals

At the heart of calculus are operations like differentiation and integration. Integration, particularly indefinite integration, computes an antiderivative of a function. This means finding a function whose derivative matches the original function. The integral sign \(\int\) denotes the operation, and the result includes an integration constant \(C\).
The process of indefinite integration is essentially reversing differentiation. For instance, if the derivative of \(F(t)\) is \(f(t)\), then \(F(t)\) is an antiderivative of \(f(t)\). The integration adds a constant \(C\) because differentiating a constant results in zero, and different functions can share the same derivative. Thus, when integrating, it's important to include \(C\) to represent all possible antiderivatives of the function.

Algebraic Integration

Algebraic integration involves finding integrals of functions defined by polynomials or algebraic expressions. It utilizes both algebraic manipulation and integration rules to simplify and integrate a given function. Looking at the original exercise, the function \(t^3 + 6t^2\) consists of polynomial terms.In algebraic integration, it is often helpful to deal with constants separately. For a term like \(6t^2\), we factor out the constant before integrating the polynomial component, thus:

• Apply \(\int 6t^2 \, dt = 6 \int t^2 \, dt\)

By managing terms with consistent logic, algebraic integration allows systematic tackling of complex integrals.
Through these methods, the indefinite integral of a polynomial like \(t^3 + 6t^2\) is computed term by term, ultimately providing a holistic result of the integrated function.