Question

Find the area of the region in the first quadrant below \(y=e^{-x}\) above \(y=\frac{1}{2}\)

Short answer

The area is \(\frac{1}{2} - \frac{1}{2}\ln(2)\).

Step-by-step solution

Sketch and Understand the Region

To find the area of the region, first sketch the graphs of the functions involved. The exponential function, \(y = e^{-x}\), is a downward-sloping curve starting from 1 when \(x = 0\) and approaching 0 as \(x\) increases. The horizontal line \(y = \frac{1}{2}\) intersects this curve in the first quadrant. We need to identify the region between these two curves.

Find the Intersection Points

To determine the boundaries of the area, find the intersection points of the two functions by setting them equal: \(e^{-x} = \frac{1}{2}\). Solve this equation to find the value of \(x\):\[ e^{-x} = \frac{1}{2} \Rightarrow x = -\ln\left(\frac{1}{2}\right) = \ln(2) \]}

Setup the Integral for the Area

The area between two curves in the first quadrant can be found by integrating the difference of the functions from the left to the right boundary.The integration limits are from \(x = 0\) to \(x = \ln(2)\). The area can be calculated as:\[ \int_{0}^{\ln(2)} (e^{-x} - \frac{1}{2}) \, dx \]

Integrate the Differential

Carry out the integration:1. Integrate \(e^{-x}\): \[ \int e^{-x} \, dx = -e^{-x} + C \]2. Integrate \(\frac{1}{2}\): \[ \int \frac{1}{2} \, dx = \frac{1}{2}x + C \]Substitute results into one integral:\[ \left[ -e^{-x} - \frac{1}{2}x \right]_{0}^{\ln(2)} \]

Evaluate the Definite Integral

Evaluate the expression at the bounds:1. For \(x = \ln(2)\): \[ -e^{- ext{ln}(2)} - \frac{1}{2}\ln(2) = -\frac{1}{2} - \frac{1}{2}\ln(2) \]2. For \(x = 0\): \[ -e^{0} - \frac{1}{2} \cdot 0 = -1 \]Subtract the lower bound evaluation from the upper:\[-\left( \frac{1}{2} + \frac{1}{2}\ln(2) \right) + 1 = \frac{1}{2} - \frac{1}{2}\ln(2) \]

Final Area Result

The calculated value of the area is thus:\[ \frac{1}{2} - \frac{1}{2}\ln(2) \]

Key concepts

Area Between Curves

When finding the area between curves, our primary goal is to find the region enclosed between two functions on a graph. This method in integral calculus helps us to understand how much space exists between curves over a given range. For this specific exercise, the area of interest lies between the curve of the exponential function, \( y = e^{-x} \), and the horizontal line \( y = \frac{1}{2} \), in the first quadrant of the Cartesian plane.

• The first quadrant is crucial because it includes only the positive x and y coordinates, helping to define the region we're interested in.
• To determine the region, visualize the graphs of both functions. The exponential curve descends from 1 while the line remains constant at \( \frac{1}{2} \).

By setting up the integral of the difference between these two functions over a specified interval, we can accurately measure the space between them.

Definite Integral

The definite integral is a powerful tool in calculus used to compute the net area under a curve from one point to another. It essentially provides a way of adding up infinitely many small vertical slices of area, computed between the x-axis and the curve over an interval. For this exercise:

• The interval between the intersection points, from \( x = 0 \) to \( x = \ln(2) \), is crucial as it forms the bounds for our integral.
• The definite integral setup is \( \int_{0}^{\ln(2)} (e^{-x} - \frac{1}{2}) \, dx \), which calculates the area by finding the difference between the two functions \( e^{-x} \) and \( \frac{1}{2} \).

To perform this calculation, solve the integral and evaluate it at both the upper and lower bounds to find the exact size of the area in question.

Exponential Function

An exponential function, like \( y = e^{-x} \), is notable for its rapid change. It's characterized by the constant base \( e \), raised to a variable exponent. Here's why it's significant in this problem:

• As \( x \) increases, the value of \( e^{-x} \) decreases towards zero, forming a classic exponential decay curve.
• Understanding this behavior helps predict the shape of the curve and its intersection with other functions, like our horizontal line.

The exponential curve continually moves close to the x-axis as \( x \) grows, which is important to visualize the area being calculated between \( y = e^{-x} \) and the straight line \( y = \frac{1}{2} \).

Intersection Points

Intersection points are where two graphs meet on a coordinate plane. These points are essential because they act as boundaries for the area that we want to calculate. In this case, finding where \( y = e^{-x} \) coincides with \( y = \frac{1}{2} \) is crucial:

• By setting the equations equal, \( e^{-x} = \frac{1}{2} \), we solve for \( x \) to find the intersection point. Here, \( x = \ln(2) \).
• This value marks the end of the interval over which we integrate our functions to find the area between them.

It's a critical step in the process because no other x-values give the same y-coordinate for both functions within the first quadrant.