Problem 27 - Applied Calculus

Question

Find each indefinite integral. \(\int\left(5 e^{0.02 t}-2 e^{0.01 t}\right) d t\)

Step-by-Step Solution

Answer

The indefinite integral is \(250 e^{0.02 t} - 200 e^{0.01 t} + C\).

Step 1: Set Up the Integral

We begin with the original integral expression given in the problem: \[ \int\left(5 e^{0.02 t}-2 e^{0.01 t}\right) d t \] The integral is a linear combination of two exponential functions.

Step 2: Break Down the Integral into Separate Terms

Using the linearity property of integrals, we can separate the integral into two separate integrals, like so: \[\int 5 e^{0.02 t} \, dt - \int 2 e^{0.01 t} \, dt. \] This allows us to handle each term individually.

Step 3: Solve the First Integral

For the first integral, \(\int 5 e^{0.02 t} \, dt\), use the integration rule for exponential functions: \[\int e^{kt} dt = \frac{1}{k}e^{kt} + C. \] Here, \(k = 0.02\), so the integral becomes: \[5 \times \frac{1}{0.02} e^{0.02 t} = 250 e^{0.02 t}. \]

Step 4: Solve the Second Integral

For the second integral, \(\int 2 e^{0.01 t} \, dt\), apply the same exponential integration rule with \(k = 0.01\). This gives: \[2 \times \frac{1}{0.01} e^{0.01 t} = 200 e^{0.01 t}. \]

Step 5: Combine the Integrals

Now that we have integrated each term, combine the results into a general solution: \[250 e^{0.02 t} - 200 e^{0.01 t} + C. \] Remember, \(C\) is the constant of integration, representing the family of indefinite integrals.

Key Concepts

Exponential Functions

Exponential functions are a fascinating topic in mathematics, often featuring variables in the exponent. They appear in a broad range of real-world contexts like population growth, radioactive decay, and compound interest.
An exponential function typically looks like this: \( f(t) = a \, e^{kt} \), where \( a \) is a constant, \( e \) is the base of the natural logarithm, and \( k \) is the rate of growth or decay.

• When \( k > 0 \), the function indicates exponential growth.
• When \( k < 0 \), it reflects exponential decay.

This characteristic makes them particularly suitable for modeling time-dependent processes. When dealing with exponential functions in calculus, it is crucial to recognize the specific form and rate \( k \) to apply integration rules effectively.

Integration Techniques

Integration techniques are essential tools for solving problems involving antiderivatives, especially with integrals that involve exponential functions. The key technique to remember when integrating exponential functions of the form \( e^{kt} \) is using the basic rule: \[ \int e^{kt} \, dt = \frac{1}{k} e^{kt} + C \]Here are useful steps to integrate effectively:

• Identify and separate exponential terms using the property of linearity in integrals, which allows breaking down larger expressions into more manageable parts.
• Apply the formula by isolating \( k \) from the exponent, multiplying the integral by \( \frac{1}{k} \).
• Ensure correct constant multiplication outside the exponential function. This could require factoring constants out of the integral expression beforehand.

Using these steps, each integral can be evaluated systematically, allowing for a clear and accurate solution.

Constant of Integration

In calculus, the constant of integration, often represented as \( C \), is a significant concept. It arises in the context of indefinite integrals, which do not have specified limits of integration. Therefore, finding an indefinite integral yields a family of functions rather than a single function.
This constant accounts for all possible vertical shifts of the antiderivative along the y-axis because differentiation of a constant is zero. When combining results after integration, always add \( C \) to ensure completeness of the solution.

• It serves as a placeholder for any constant that could have been part of the original function before differentiation.
• In application problems, if additional conditions are supplied (such as initial conditions), they can be used to determine the specific value of \( C \).

Remember, omitting the constant of integration can lead to incomplete solutions in indefinite integration problems.