Problem 77 - Calculus and its applications

Question

Solve each integral. Each can be found using rules developed in this section, but some algebra may be required. $$ \int(5 t+4)^{2} t^{4} d t $$

Step-by-Step Solution

Answer

\( \frac{25}{7}t^7 + \frac{20}{3}t^6 + \frac{16}{5}t^5 + C \)

Step 1: Expand the Integrand

To solve the integral \( \int (5t + 4)^2 t^4 \, dt \), first expand \((5t + 4)^2\) into \(25t^2 + 40t + 16\). This gives us:\[ \int (25t^2 + 40t + 16) t^4 \, dt = \int (25t^6 + 40t^5 + 16t^4) \, dt \]

Step 2: Split the Integral

Next, split the integral into three separate integrals:\[ \int 25t^6 \, dt + \int 40t^5 \, dt + \int 16t^4 \, dt \]

Step 3: Apply the Power Rule

Use the power rule for integration, which states \( \int t^n \, dt = \frac{t^{n+1}}{n+1} + C \) for each term:1. \( \int 25t^6 \, dt = \frac{25}{7}t^7 \)2. \( \int 40t^5 \, dt = \frac{40}{6}t^6 \)3. \( \int 16t^4 \, dt = \frac{16}{5}t^5 \)

Step 4: Simplify the Results

Simplify each term and combine the results:1. \( \frac{25}{7}t^7 2. + \frac{20}{3}t^6 3. + \frac{16}{5}t^5 \)The integrated expression becomes:\[ \frac{25}{7}t^7 + \frac{20}{3}t^6 + \frac{16}{5}t^5 + C \]

Step 5: Compilation of the Final Answer

Combine the simplified terms into the final solution. Don't forget the constant of integration \( C \):\[ \int (5t + 4)^2 t^4 \, dt = \frac{25}{7}t^7 + \frac{20}{3}t^6 + \frac{16}{5}t^5 + C \]

Key Concepts

Integration Techniques

Integration techniques are essential tools in calculus that help us find the antiderivatives or integrals of functions. One common technique involves breaking down a complicated expression into simpler parts so it can be handled easily. In the given problem, we need to integrate the function \[\int (5t + 4)^2 t^4 \, dt\]by expanding and simplifying it.
Here’s a friendly tip: if you see a complex expression like \((5t + 4)^2\), it may help to expand it using algebraic formulas.
For this exercise, the expression was expanded to \((25t^2 + 40t + 16)\).Once it's expanded, it multiplies with \(t^4\), leading to the need to split the integral into smaller, manageable parts like:

• \(\int 25t^6 \, dt\)
• \(\int 40t^5 \, dt\)
• \(\int 16t^4 \, dt\)

These components are simpler to integrate individually, ensuring the overall process becomes systematic and less overwhelming.

Power Rule for Integration

The power rule for integration is one of the simplest and most frequently used techniques in integral calculus. It states that the integral of \(t^n\) is \(\frac{t^{n+1}}{n+1} + C\),where \(neq -1\), and \(C\)is the constant of integration.
In our example, after breaking down the integral into smaller parts, each individual term was processed using the power rule. Here’s how it worked for each term:

• For \(\int 25t^6 \, dt\), we applied the power rule: \[ \frac{25}{7}t^7 \]
• For \(\int 40t^5 \, dt\), it became: \[ \frac{40}{6}t^6 \]
• Finally, for \(\int 16t^4 \, dt\), it turned into: \[ \frac{16}{5}t^5 \]

By systematically applying the power rule to each term, we broke down a complex problem into a series of straightforward calculations. This approach not only simplifies integration tasks but also strengthens understanding of how different components of polynomial expressions interact when integrated.

Polynomial Integration

Polynomial integration is essentially the application of the integration techniques, specifically the power rule, to functions that are polynomials. Polynomials are expressions made up of variables and coefficients, consisting of terms in the form of \(a_n t^n\).In our exercise, once we expanded \((5t + 4)^2 t^4\)into a polynomial with terms like \(25t^6\), \(40t^5\), and \(16t^4\), we applied integration to each term individually.
The beauty of polynomial integration is its straightforwardness once the polynomial is broken down term by term.

• The term \(25t^6\)integrates smoothly to \(\frac{25}{7}t^7\),
• \(40t^5\)advances to \(\frac{20}{3}t^6\),
• and \(16t^4\) simplifies to \(\frac{16}{5}t^5\).

Combining these individual results gives us the integral of the polynomial, plus the constant \(C\).Polynomial integration is not just about tearing complex expressions apart; it's about gaining insights into the underlying structure of equations, making them manageable in bite-sized chunks. Once you master this technique, polynomial integrals become as friendly as they are fundamental in calculus.