Problem 28 - Applied Calculus

Question

Find each indefinite integral. \(\int\left(3 e^{0.05 t}-2 e^{0.04 t}\right) d t\)

Step-by-Step Solution

Answer

The indefinite integral is \( 60 e^{0.05t} - 50 e^{0.04t} + C \).

Step 1: Break Down the Integral

The integral is given as \( \int (3e^{0.05t} - 2e^{0.04t}) \ dt \). Recognizing that integration is linear, we can separate this into two separate integrals: \( \int 3e^{0.05t} \ dt \) and \( \int -2e^{0.04t} \ dt \).

Step 2: Integrate Each Part Separately

To integrate \( \int 3e^{0.05t} \ dt \), use the formula \( \int ae^{bt} \ dt = \frac{a}{b}e^{bt} + C \). Here, \( a = 3 \) and \( b = 0.05 \), so the integral becomes \( \frac{3}{0.05} e^{0.05t} = 60 e^{0.05t} \).

Step 3: Integrate the Second Term

Now, integrate \( \int -2e^{0.04t} \ dt \) using the same formula. Here, \( a = -2 \) and \( b = 0.04 \), resulting in \( \frac{-2}{0.04} e^{0.04t} = -50 e^{0.04t} \).

Step 4: Combine Results and Add Constant

Combine the results from Steps 2 and 3: \( 60 e^{0.05t} - 50 e^{0.04t} \). Since it's an indefinite integral, remember to add the constant of integration \( C \). Hence, the final solution is \( 60 e^{0.05t} - 50 e^{0.04t} + C \).

Key Concepts

Exponential Function Integration

When integrating exponential functions, we often encounter expressions of the form \( a e^{bt} \). To integrate such expressions, we apply a basic formula: \( \int a e^{bt} \ dt = \frac{a}{b} e^{bt} + C \). This formula is derived from recognizing the derivative of an exponential function is the function itself multiplied by its exponent's coefficient.
In our exercise, we see two parts: \( 3e^{0.05t} \) and \( -2e^{0.04t} \). Breaking it down:

• For \( 3e^{0.05t} \), where \( a = 3 \) and \( b = 0.05 \), the integration results in \( \frac{3}{0.05} e^{0.05t} = 60 e^{0.05t} \).
• For \( -2e^{0.04t} \), where \( a = -2 \) and \( b = 0.04 \), it results in \( \frac{-2}{0.04} e^{0.04t} = -50 e^{0.04t} \).

The linearity of integration allows us to sum these results, as explained in calculus problem-solving.

Calculus Problem Solving

Solving calculus problems, such as integrating complex expressions, requires multiple steps. Begin by understanding the properties of integration, like linearity. This allows decomposing integrals of multiple terms into simpler parts to tackle individually. Once separated, each piece is handled using relevant integration formulas or techniques.
In our specific exercise:

• The original integral \( \int (3e^{0.05t} - 2e^{0.04t}) \ dt \) was divided into two separate integrals: \( \int 3e^{0.05t} \ dt \) and \( \int -2e^{0.04t} \ dt \).
• Applying the known exponential integral formula to each part is straightforward, aiding in obtaining accurate results efficiently.
• A final check ensures all iterations and assumptions align, especially with constants and coefficients.

Each result is then combined to yield the integrated expression.

Constant of Integration

The constant of integration, denoted as \( C \), plays a crucial role in indefinite integrals. It represents the family of all antiderivatives for a given function. Since the differentiation of a constant is zero, adding \( C \) accounts for all possible vertical shifts of the antiderivative function.
Whenever performing indefinite integration:

• Always append \( C \) to the integral result, capturing the general solution set.
• Remember that without \( C \), the answer might seem incomplete, omitting part of the solution family.

In our exercise, we conclude with \( 60 e^{0.05t} - 50 e^{0.04t} + C \), ensuring all potential solutions related to different values of \( C \) are considered. Thus, this inclusion is invaluable for capturing the entirety of indefinite integrals' solutions.