Problem 2 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

Check by differentiation the validity of the indefinite integral formulas: a. \(\int \frac{1}{t} \mathrm{dt}=\ln \mathrm{t}+\mathrm{C}\) b. \(\int[\mathrm{U}(\mathrm{t})]^{\mathrm{n}} \mathrm{U}^{\prime}(\mathrm{t}) \mathrm{dt}=\frac{\mathrm{U}(\mathrm{t})^{\mathrm{n}+1}}{\mathrm{n}+1}+\mathrm{C}\) c. \(\int \mathrm{e}^{\mathrm{kt}} \mathrm{dt}=\frac{1}{\mathrm{k}} \mathrm{e}^{\mathrm{kt}}+\mathrm{C}\)

Step-by-Step Solution

Answer

All the indefinite integral formulas are valid, as differentiation yields the original integrands.

Step 1: Introduction

To verify the validity of indefinite integral formulas, we use differentiation. If differentiating the supposed integral yields the original integrand, the formula is valid.

Step 1: Differentiate Part (a)

Differentiate \ \(\ln(t) + C\) \ with respect to \(t\). The derivative of \(\ln(t)\) is \(\frac{1}{t}\), and the derivative of \(C\), a constant, is 0. Thus, differentiating gives \(\frac{1}{t}\), which matches the original integrand \(\frac{1}{t}\).

Step 2: Differentiate Part (b)

Apply the chain rule to differentiate \ \(\frac{U(t)^{n+1}}{n+1} + C\) \ with respect to \(t\). \ The derivative is \(U(t)^n \cdot U'(t)\), exactly the original integrand \ ([U(t)]^{n} \cdot U'(t)) \.

Step 3: Differentiate Part (c)

Differentiate \ \(\frac{1}{k}e^{kt} + C\) \ with respect to \(t\). Using the chain rule, the derivative is \ \(e^{kt}\), matching the original integrand \(e^{kt}\).

Step 5: Conclusion

By differentiation, all three integrals' derivatives produce their respective original integrands, confirming their validity.

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus. It involves finding the rate at which a function changes at any given point. When you see the term "differentiate," it means you should find the derivative of a function. Differentiation helps us verify indefinite integrals by reversing the process of integration, essentially "undoing" it. For instance, if we have a function like \(f(t) = \ln(t) + C\), differentiating it with respect to \(t\) involves using the derivative rules for logarithms.

• The derivative of \(\ln(t)\) is \(\frac{1}{t}\).
• The derivative of any constant \(C\) is 0.

By differentiating the result of an indefinite integral, you can check if you get the original function inside the integral sign. If both match, then the integral formula is correct.
Understanding differentiation helps in mastering various calculus concepts such as curve sketching, optimizing functions, and working with series and sequences. It's a tool that forms the foundation for solving many real-world problems in physics, engineering, and economics.

Verification of Integrals

Verification of an integral's validity is done by differentiating the result of the indefinite integral. The goal is to determine whether the differentiation yields the original integrand. This method ensures that the integral has been calculated correctly.
For example, in the exercise above:

• Part (a): Differentiating \(\ln(t) + C\) gives \(\frac{1}{t}\), matching the original integrand.
• Part (b): The derivative of \(\frac{U(t)^{n+1}}{n+1} + C\) is \(U(t)^n \cdot U'(t)\), the original function.
• Part (c): Differentiating \(\frac{1}{k}e^{kt} + C\) gives \(e^{kt}\), the starting expression inside the integral.

Each step confirms that differentiating the result returns the original expression we want to integrate, verifying the formula's correctness.
Verification is a crucial practice not only for textbook exercises but also in ensuring the solution to real-world problems is accurate and reliable. It's a bridging step between integration and differentiation, tying these processes together for consistent results.

Chain Rule

The Chain Rule is a technique in calculus used for differentiating compositions of functions. It allows us to find derivatives of complex expressions that involve a function inside another function. The chain rule states that the derivative of \(f(g(t))\) is \(f'(g(t)) \, \cdot \, g'(t)\).
In our exercise, the chain rule comes into play particularly in Part (b) and Part (c):

• For \(\frac{U(t)^{n+1}}{n+1} + C\), it helps us find the derivative as \(U(t)^n \cdot U'(t)\). Here, \(U(t)\) is the inner function, and \(U'(t)\) is its derivative.
• In Part (c), when differentiating \(\frac{1}{k}e^{kt} + C\), the chain rule ensures we account for the rate of change inside the exponential, leading us to the correct derivative \(e^{kt}\).

Utilizing the chain rule allows for precise calculations of derivatives, ensuring accuracy in composing different functions into one. Understanding and applying the chain rule is vital for complex calculus problems and is extensively used in higher mathematics and various scientific disciplines.