Problem 23 - Applied Calculus

Question

Find each indefinite integral. \(\int\left(3 e^{0.5 t}-2 t^{-1}\right) d t\)

Step-by-Step Solution

Answer

Integral: \( 6 e^{0.5t} - 2 \ln |t| + C \).

Step 1: Break Down the Integral

The given integral is \( \int \left(3 e^{0.5 t} - 2 t^{-1}\right) \, dt \). We can separate this into the sum of two integrals: \( \int 3 e^{0.5 t} \, dt - \int 2 t^{-1} \, dt \).

Step 2: Integral of the Exponential Function

Let's solve the first integral \( \int 3 e^{0.5 t} \, dt \). We use the substitution method. Set \( u = 0.5t \), then \( du = 0.5 \, dt \) or \( dt = \frac{2}{1} du \). Substitute into the integral: \( 3 \int e^u \, \frac{2}{1} du = 6 \int e^u \, du = 6e^u + C_1 \). Substituting back for \( u \), we get \( 6 e^{0.5t} + C_1 \).

Step 3: Integral of the Reciprocal Function

Now solve the second integral \( \int 2 t^{-1} \, dt \), which is a standard integral of the form \( \int \frac{2}{t} \, dt = 2 \ln |t| + C_2 \).

Step 4: Combine the Integrals

Add the results of the two integrals from Step 2 and Step 3: \( 6 e^{0.5t} - 2 \ln |t| + C \), where \( C = C_1 - C_2 \) is the overall constant of integration.

Key Concepts

Exponential Function

In calculus, one of the most important functions is the exponential function, often written as \(e^x\). This function grows rapidly and appears frequently in real-world applications such as compound interest and population modeling. In our problem, we have an exponential function \(3e^{0.5t}\), which means that the rate of growth, denoted by \(0.5t\), is multiplied by 3. Understanding this allows us to address integrals involving exponential functions effectively.

When integrating exponential functions, we rely heavily on the inherent properties of the exponential. Typically, an expression \(e^{kt}\) integrates to \(\frac{1}{k} e^{kt} + C\), where \(k\) is a constant, and \(C\) is an integration constant. Recognizing the rate \(k\) in our function is vital for correct application, as this multiplier adjusts the rule accordingly.

In the provided solution, to tackle the integral of \(3e^{0.5t}\), the substitution method helps in simplifying expressions with exponential derivatives, which leads us to a seamless integration process.

Substitution Method

The substitution method is a powerful tool in calculus for solving complex integrals more easily by simplifying them into an easier form. This method is especially useful when faces with compositions of functions, such as products of functions or exponential functions raised to a power.

For the integral \(\int 3 e^{0.5 t} \, dt\), substitution simplifies the calculation process. The idea is to substitute \(u = 0.5t\), which means that the differential \(dt\) must be replaced accordingly. By expressing \(dt\) in terms of \(du\), as \(dt = \frac{2}{1} du\), the integral becomes more straightforward:

• Substitute \(u = 0.5t\), making it \(\int 3 e^u \times \frac{2}{1} du\)
• The integral simplifies to \(6 \int e^u du = 6e^u + C_1\)
• Finally, revert substitution: replace \(u\) with \(0.5t\) to get \(6e^{0.5t} + C_1\)

The method's beauty lies in simplifying complex functions into linear ones, thus easing the integration process and revealing the solution directly.

Reciprocal Function

Another essential concept in calculus is the reciprocal function, represented as \(t^{-1}\) in the integral \(\int 2t^{-1} \, dt\). These functions are inversely related to linear functions and have distinctive integration properties that differ from polynomial functions.

The indefinite integral of a reciprocal function \(\int t^{-1} \, dt\) is a well-known result: \(\ln|t| + C\), where \(C\) is the integration constant. This stems from the derivative of the natural logarithm function being \(\frac{1}{t}\).

Incorporating this into our exercises, we examine \(\int 2 t^{-1} \, dt\), which simplifies to:

• Using the integral property: \(2 \int t^{-1} \, dt = 2 \ln |t| + C_2\)

The reciprocal's integration admits a natural log function, which serves to counterbalance the original function's inverse relation. Pairing this result with the exponential integral, we arrive at a comprehensive solution for the entire indefinite integral.