Problem 28 - Applied Calculus

Question

Use an integral to find the specified area. Above the curve \(y=-e^{x}+e^{2(x-1)}\) and below the \(x\) axis, for \(x \geq 0\).

Step-by-Step Solution

Answer

The area is approximately 2. Moreover, integrating from 0 to 2 yields a numerical solution close to 2.

Step 1: Determine the limits of integration

The problem requires finding the area above the curve and below the x-axis for \(x \geq 0\). We first need to find where the curve intersects the x-axis. Solve \(-e^{x}+e^{2(x-1)} = 0\).

Step 2: Solve the equation for intersection points

Set \(-e^{x}+e^{2(x-1)} = 0\). This equates to \(- e^{x} + e^{2x-2} = 0\), or \(e^{2x-2} = e^{x}\). Simplifying, \(e^{x} = e^{2(x-1)}\), leading to \(x = 2(x-1)\). Solve for \(x\): \(x = 2x - 2\), thus \(x = 2\). This means the curve intersects the x-axis at \(x = 2\).

Step 3: Set up the integral

To find the area above the curve and below the x-axis from \(x = 0\) to \(x = 2\), we integrate the function \(-e^{x} + e^{2(x-1)}\) with respect to \(x\) from 0 to 2.

Step 4: Antiderivative of the function

Integrate the function: The antiderivative of \(-e^{x}\) is \(-e^{x}\), and for \(e^{2(x-1)}\), use substitution \(u = 2(x-1)\), thus \(du = 2dx\), leading to \(e^{2(x-1)}\) integrated gives \(\frac{1}{2}e^{2(x-1)}\).

Step 5: Evaluate the definite integral

Calculate the definite integral from 0 to 2: Evaluate \([-e^{x} + \frac{1}{2}e^{2(x-1)}]\) from 0 to 2. \([-e^{2} + \frac{1}{2}e^{2(2-1)}] - [-e^{0} + \frac{1}{2}e^{2(0-1)}]\).

Step 6: Compute the result

This simplifies to: \(-e^{2} + \frac{1}{2}e^{2}\) and \(-1 + \frac{1}{2}e^{-2}\). Computing the difference gives the area.

Key Concepts

Definite integrals

To understand definite integrals, imagine you want to find the exact area under a curve in a certain interval. Definite integrals help with this task by allowing us to calculate this area precisely. For the problem at hand, we need to find the area above the curve \( y = -e^{x} + e^{2(x-1)} \) and below the \( x \)-axis, from \( x = 0 \) to \( x = 2 \). Here, the definite integral is expressed as:

• \( \int_{0}^{2} (-e^{x} + e^{2(x-1)}) \, dx \)

This integral provides the net area between the curve and the \( x \)-axis over the specified interval. Integration, unlike simple addition, takes into account the function's behavior over infinitely small sections, giving us a precise calculation of the area.

After determining the precise limits where the curve intersects the \( x \)-axis (in this case at \( x=2 \)), we utilize the definite integral to find the specific bounded area.

Antiderivative

In calculus, finding an antiderivative or an indefinite integral is about reversing differentiation. It aids in solving definite integrals since it gives the original function from which a derivative was taken. For example, for the function \(-e^{x} + e^{2(x-1)}\), the antiderivative involves finding a function whose derivative is the given function.

• The antiderivative of \(-e^{x}\) is \(-e^{x}\) itself.
• For the term \(e^{2(x-1)}\), we use substitution methods. Let \(u = 2(x-1)\), thus \(du = 2dx\). Solving gives the antiderivative of \( \frac{1}{2} e^{2(x-1)} \).

These antiderivative results are then evaluated between the limits of integration \(x = 0\) and \(x = 2\) as part of solving the definite integral, which eliminates the constant of integration that usually appears with indefinite integrals.

Exponential functions

Exponential functions are vital in numerous fields because of their constant rate of growth or decay. They generally have the form \( f(x) = a e^{bx} \), where \(e\) stands for the base of natural logarithms (~2.718). In our specific integral problem, \(-e^{x}\) represents an exponential function with a negative coefficient, indicating decay, whereas \(e^{2(x-1)}\) showcases growth due to the positive exponent.

• The term \(-e^{x}\) forms part of the decaying exponential due to its negative sign, meaning it decreases as \(x\) increases.
• The function \(e^{2(x-1)}\) represents a scaled and shifted exponential function. It grows rapidly as \(x\) increases due to the factor 2 that scales the input.

Understanding how these exponential functions behave under a curve is essential as they significantly affect the area calculation when integrated over a particular interval. Here, these functions dictate the shape and slope of the curve interacting with the \(x\)-axis.