91Ó°ÊÓ

Exponential functions are a critical component in calculus and mathematics as a whole. They are functions in which the variable appears in the exponent. A simple example is the base of natural logarithms, represented as \(e^x\). These functions have unique properties:

• Their derivative is proportional to the original function.
• They grow (or decay, if negative) continuously and rapidly.

In the context of calculus, understanding exponential functions is vital because they model many real-world processes like population growth and radioactive decay. In our problem, we encounter the exponential function \(e^{-0.2t}\). This particular function represents exponential decay at a rate of 0.2. Comprehending these properties aids in integrating such functions, ensuring the work connects naturally to real scenarios requiring calculus.

Integration techniques are the methods used to find integrals of functions. The integral \(\int 100 e^{-0.2 t} d t\) can be solved using a basic integration rule for exponential functions. To integrate \(e^{kt}\), you apply the formula \( \int e^{kt} \, dt = \frac{1}{k} e^{kt} + C \), where \(k\) is a constant and \(C\) represents the constant of integration.In our specific exercise:

• Identify \(k = -0.2\) from the exponent.
• Use the rule: \( \int e^{-0.2 t} \, dt = \frac{1}{-0.2} e^{-0.2 t} + C \).
• Multiply by 100 (a constant factor outside the integral) yielding the result: \(-500 e^{-0.2 t} + C\).

This integration process shows how applying the correct techniques systematically helps solve complex problems.

Differentiation and integration are core operations in calculus that are inverse processes. Differentiation refers to finding the rate at which a quantity changes, while integration involves finding a quantity given its rate of change.To verify the result of an integration, differentiation is often used to "check the answer." For this exercise:

• Differentiating \(-500 e^{-0.2 t} + C\) with respect to \(t\).
• Applying derivative operation: \( \frac{d}{dt}(-500 e^{-0.2 t} + C) = -500 \cdot (-0.2) e^{-0.2 t}\).
• Resulting in the original function: \(100 e^{-0.2 t}\).

This step ensures accuracy and correctness, showing the true value of using both integration and differentiation in tandem, strengthening understanding of their relationships in solving real-life mathematical problems.