Problem 12 - Applied Calculus

Question

Find each indefinite integral. \(\int 24 e^{-2 u / 3} d u\)

Step-by-Step Solution

Answer

The indefinite integral is \(-36 e^{-2u/3} + C\).

Step 1: Identify the Exponential Function

The integrand is given as \(24 e^{-2u/3}\). Notice that this is an exponential function with respect to \(u\).

Step 2: Apply the Integration Formula

Use the formula \(\int e^{ax} \, dx = \frac{1}{a} e^{ax} + C\). In this case, \(a = -\frac{2}{3}\).

Step 3: Simplify the Integral

Substitute \(a = -\frac{2}{3}\) into the formula: \(\int 24 e^{-2u/3} \, du = 24 \cdot \left(-\frac{3}{2}\right) e^{-2u/3} + C\).

Step 4: Perform the Multiplication

Calculate the constant multiplier: \(24 \cdot \left(-\frac{3}{2}\right) = -36\).

Step 5: Write the Final Integral

Thus, the integral becomes \(-36 e^{-2u/3} + C\), where \(C\) is the constant of integration.

Key Concepts

Exponential Functions

Exponential functions are central in calculus since they undergo rapid changes. They have the general form \( e^{x} \), where \( e \) is Euler's number, approximately 2.71828. In our exercise, the function is \( e^{-2u/3} \). In this example, the exponent is a linear function of \( u \), specifically \(-\frac{2}{3}u \).

• Such functions are very useful in modeling growth and decay processes.
• The constant \(-\frac{2}{3}\) serves as a rate or scale factor for \( u \).

Recognizing exponential functions will help us in determining which integration techniques to apply. Their unique properties make them straightforward to integrate when the exponent is linear.

Integration Techniques

Integration techniques allow us to find the antiderivative of a function. In the case of exponential functions, we use the formula \( \int e^{ax} \, dx = \frac{1}{a} e^{ax} + C \), where \( a \) is a constant. This formula is deduced from the fact that the derivative of \( e^{ax} \) gives us \( ae^{ax} \).

• Identify the constant \( a \): This is crucial, as integrating involves dividing the function by this constant.
• For our problem, substituting \( a = -\frac{2}{3} \), we modify the formula to accommodate this rate of change.

By manipulating the constants and keeping the variable structure, the indefinite integral can be evaluated swiftly. In our example, multiplication of constants was necessary after applying the formula.

Calculus Problem Solving

Solving calculus problems often involves a step-by-step logical approach. This ensures each part contributes effectively to the solution.

• Start by identifying the type of function you're dealing with—useful for applying the right integration technique.
• Substitute values attentively, especially for constants like \( a \) in exponential functions.
• Simplify the expressions and perform necessary arithmetic operations.

In the given exercise, recognizing the negative rate \(-\frac{2}{3}\), multiplying constants correctly, and ensuring every part is integrated is vital. Calculus is all about accurate and efficient steps leading to a meaningful solution, so practice is key to mastering these techniques.