Problem 5 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Find the area under the graph of \(y=e^{x}\) between \(x=0\) and \(x=1\).

Step-by-Step Solution

Answer

The area under the graph of \(y=e^{x}\) between \(x=0\) and \(x=1\) is \(e - 1\).

Step 1: Write down the function

The function given in the exercise is \(y=e^{x}\).

Step 2: Set up the definite integral

The definite integral will be from 0 to 1. So, set it up as \(\int_{0}^{1} e^{x} dx\).

Step 3: Evaluate the integral

To evaluate the integral, you need perform the antiderivatives of \(e^{x}\). The antiderivative of \(e^{x}\) is simply \(e^{x}\). So, the evaluation of the integral becomes \(e^{x}\Bigg|_0^1\).

Step 4: Subtract the lower limit value from the upper limit value

First evaluate for the upper limit 1, \(e^{1} = e\), then for the lower limit 0, \(e^{0} = 1\). Now, subtract the lower limit value from the upper limit which gives \(e - 1\).

Key Concepts

Exponential Functions

Exponential functions are fundamental in understanding growth and decay processes in mathematics. In the equation given, we have the function \(y=e^{x}\), which is one of the simplest forms of exponential functions. The base of this function is the mathematical constant \(e\), approximated as 2.71828. Exponential functions have the unique property of increasing swiftly as the variable \(x\) increases.

This rapid growth is due to the fact that the rate of change of \(e^{x}\) with respect to \(x\) is proportional to its current value. Such functions are prevalent in many real-world contexts, like in financial calculations for compound interest, population growth models, and natural decay processes. Understanding exponential functions is crucial as they serve as models for a broad spectrum of phenomena.

If you visualize the graph of \(e^{x}\), it rapidly rises as you move to the right along the x-axis. This characteristic makes integrating exponential functions interesting, especially when determining the area under such curves, which we'll explore through definite integrals.

Antiderivatives

Antiderivatives, also known as indefinite integrals, are crucial for solving calculus problems involving differentiation and integration. An antiderivative is essentially the reverse process of differentiation. If you start with a function and differentiate it, finding the antiderivative will take you back to the original function.

For exponential functions like \(e^{x}\), this process is peculiarly straightforward. The antiderivative of \(e^{x}\) remains \(e^{x}\). This is due to the unique property of \(e^{x}\) which states that its derivative and its integral are the same.

When we integrate \(e^{x}\) between two points, we know instantly that the indefinite integral remains \(e^{x}\). To solve the definite integral, which helps us find the area under a curve, evaluating the antiderivative at given bounds provides the solution. This leads us directly to the final result without the need for complex transformations or additional constants. Thus, understanding antiderivatives is key for solving definite integrals effortlessly.

Calculus Problem Solving

Calculus problem solving involves structured approaches to analyze and solve problems related to change. In the context of finding areas under curves, calculus uses the integral operation extensively.

In our specific problem, we are tasked with finding the area under the curve \(y=e^{x}\) from \(x=0\) to \(x=1\). This task requires setting up a definite integral: \(\int_{0}^{1} e^{x} dx\).

Here's a concise way to approach such problems:

• Identify the function and determine the appropriate limits of integration.
• Calculate the antiderivative of the function. For \(e^{x}\), the antiderivative is simply \(e^{x}\).
• Evaluate the antiderivative at the upper limit and subtract the value at the lower limit.

This step-by-step method allows us to evaluate the integral, effectively providing the area under the curve. In this scenario, evaluating \(e^{x}\) from 0 to 1 gives us \(e-1\). This process illustrates how calculus problem solving not only provides solutions but also deepens our comprehension of mathematical relationships and their applications in various fields.