Problem 12 - Calculus Volume 1

Question

For the following exercises, graph the equations and shade the area of the region between the curves. Determine its area by integrating over the \(x\) -axis. $$ y=e, y=e^{x}, \text { and } y=e^{-x} $$

Step-by-Step Solution

Answer

The area between the curves is \(2e\).

Step 1: Identify the intersection points

The curves intersect at points where \(y = e^x\) and \(y = e^{-x}\) are equal to \(y = e\). To find these points, set \(e^x = e\) and \(e^{-x} = e\). Solving these gives \(x = 1\) and \(x = -1\), respectively.

Step 2: Set up the integral to find the area

The area bounded by the curves is between \(x = -1\) and \(x = 1\). The upper curve is \(y = e\), while the lower curves are \(y=e^{x}\) from \(x = -1\) to \(x = 0\), and \(y = e^{-x}\) from \(x = 0\) to \(x = 1\). Therefore, the integral to find the area is split into two parts: \[ A = \int_{-1}^{0} (e - e^x)\,dx + \int_{0}^{1} (e - e^{-x})\,dx. \]

Step 3: Integrate the first part

Integrate \(\int_{-1}^{0} (e - e^x)\,dx\). This can be rewritten and solved as follows:\[ A_1 = \int_{-1}^{0} e \, dx - \int_{-1}^{0} e^x \, dx = e \cdot [x]_{-1}^{0} - [e^x]_{-1}^{0}. \] Calculate the definite integrals to get:\[ A_1 = e\cdot(0 - (-1)) - (1 - \frac{1}{e}) = e + \frac{1}{e} - 1. \]

Step 4: Integrate the second part

Integrate \(\int_{0}^{1} (e - e^{-x})\,dx\). This can be rewritten and solved as follows:\[ A_2 = \int_{0}^{1} e \, dx - \int_{0}^{1} e^{-x} \, dx = e \cdot [x]_{0}^{1} - [-e^{-x}]_{0}^{1}. \] Calculate the definite integrals to get:\[ A_2 = e \cdot (1 - 0) - (\frac{1}{e} - (-1)) = e - \frac{1}{e} + 1. \]

Step 5: Sum the areas from both integrals

Add the areas found in Steps 3 and 4 to determine the total bounded area:\[ A = A_1 + A_2 = (e + \frac{1}{e} - 1) + (e - \frac{1}{e} + 1). \] Simplifying the expression gives:\[ A = 2e. \]

Key Concepts

Integral Calculus

Integral calculus helps us find the area between curves. It is involved in calculating areas, volumes, central points, and many other useful things. When finding the area, the integral can act like a process of adding up many tiny parts. Each of these parts is a slim vertical slice of the space between the curves.
To find the area between curves specifically, we calculate the integral of their difference. This essentially sums up the gap between the two functions over the specified range.

• Integral calculus is key in calculating areas.
• We sum the difference between curves to find the area in between.

Integrals are like the "opposite" of derivatives. While derivatives measure change (e.g., how fast a curve rises or falls), integrals add up small pieces to reveal how much space an object covers.

Definite Integrals

Definite integrals allow us to compute the exact size of a space, and they have a start and an end point along the horizontal axis. In this problem, we use definite integrals to calculate the area between the graphs of the functions.
When we evaluate a definite integral, it provides the total area between the curve and the x-axis over the interval, taking heights from a lower function to an upper one.

• Definite integrals have specific starting and ending points.
• They calculate the total area between function graphs over a given range.

In simpler terms, a definite integral is a way of finding how much area is squished between a graph's line and the axis, from one x-value to another.

Function Intersections

Function intersections are where curves cross each other. Finding these points is crucial because they define the range over which we integrate to find the area between the functions.
To find intersections, we set the equations equal and solve for their common values of x. This often involves basic algebra and might require some substitution or simplification.

• Intersections mark the limits of integration.
• We solve for x to find where curves intersect.

In the example, the curves intersect at points where the functions equal a specific value, guiding us to the integration boundaries from x = -1 to x = 1.