Problem 11 - Applied Calculus

Question

Find the integrals .Check your answers by differentiation. $$\int 5 e^{5 t+2} d t$$

Step-by-Step Solution

Answer

The integral is \(e^{5t+2} + C\).

Step 1: Recognize the function to be integrated

We need to integrate the function \(5 e^{5t + 2}\) with respect to \(t\). This is an exponential function where the exponent is linear in terms of \(t\).

Step 2: Set up the integration by substitution

Use the substitution method to simplify the integration. Let \(u = 5t + 2\), then \(\frac{du}{dt} = 5\), so \(dt = \frac{du}{5}\). We will substitute these into the integral.

Step 3: Substitute and simplify the integral

Substitute \(u\) and the corresponding \(dt\) into the integral: \[ \int 5 e^{u} \cdot \frac{du}{5} = \int e^{u} \, du \]Now, the integral simplifies to the basic form \(\int e^{u} \, du\).

Step 4: Integrate the simplified function

The integral of \(e^u\) is simply \(e^u + C\), where \(C\) is the constant of integration. Thus, \[ \int e^u \, du = e^u + C \]

Step 5: Substitute back to original variable

Substitute back the expression for \(u\) in terms of \(t\): \[ e^{u} = e^{5t + 2} \]Thus, the integral becomes: \[ e^{5t+2} + C \]

Step 6: Confirm by differentiation

Differentiate the result \(e^{5t+2} + C\) with respect to \(t\): \[ \frac{d}{dt}[e^{5t+2} + C] = 5e^{5t+2} \]This is the original function, confirming the integral was computed correctly.

Key Concepts

Understanding Exponential Functions

Exponential functions are a fundamental part of calculus and mathematics as a whole. Generally, they take the form of \( e^{ax+b} \), where \( e \) is the base of the natural logarithm, and \( a \) and \( b \) are constants.
An essential property of exponential functions is their rate of growth or decay, which is proportional to their current value. Simply put, the larger the value of the function, the faster it grows or decays.

• Recognizing these functions is crucial when setting up integration or differentiation problems.
• They often appear in problems involving continuous growth or decay, such as population models or radioactive decay.

In the given exercise, the function to be integrated, \( 5e^{5t + 2} \), is an exponential function with a linear exponent \( 5t + 2 \).
This means the rate of change of the function is itself a function of how large it currently is. When integrating such functions, the first step is usually to reduce the complexity using techniques like substitution.

Integration by Substitution Simplified

Integration by substitution is a method used to simplify complex integrals, particularly useful for exponential functions of a linear term.
The basic idea is to substitute the complex part of the integral with a single variable, often \( u \), which transforms the problem into a more manageable form.

• Begin by choosing a substitution that will simplify the integral.
• For \( \int 5e^{5t+2} \, dt \), we identified \( u = 5t + 2 \).
• Calculate \( \frac{du}{dt} \), which in this case is 5, allowing us to isolate \( dt \) as \( dt = \frac{du}{5} \).

After substituting, the integral simplifies to \( \int e^u \, du \). This transforms an initially intimidating-looking integral into a straightforward one.
Once you have the simplified integral, you can integrate the function as usual, resulting in \( e^u + C \).
The final step is to substitute back the original terms, so \( u = 5t + 2 \), making the integral result \( e^{5t+2} + C \).

Differentiation to Check Integration

After performing an integration, it's always a reliable practice to check your work by differentiating the result. Differentiating the integrated function should give you back the original function you started with if your computations were correct.
Here's how it works:

• Take the derived function \( e^{5t+2} + C \).
• Differentiate it with respect to \( t \).
• The result should match the function you originally integrated if all steps were properly executed.

In this case, differentiating \( e^{5t+2} \) with respect to \( t \) yields \( 5e^{5t+2} \). The constant \( C \) disappears during differentiation because the derivative of a constant is zero.
This confirms the integration correctly recaptures the original function \( 5e^{5t+2} \). Thus, by using differentiation, you reinforce the accuracy of your integration, providing a strong technique for self-verification in calculus problems.
Remember, differentiation is like a litmus test for integration reliability, a quick check to ensure the correctness of your solution.