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Sketch the region and find its area. The region bounded by \(y=e^{x}, y=e^{-2 x},\) and \(x=\ln 4\)

Short Answer

Expert verified
#tag_shortanswer#The area of the region bounded by the given curves is \(\frac{1}{2}\).

Step by step solution

01

Find intersections of \(y = e^x\) and \(y = e^{-2x}\)

To find the point(s) of intersection of the curves, we need to set \(y = e^x\) equal to \(e^{-2x}\) and solve for x: $$e^x = e^{-2x}$$ Divide both sides by \(e^x\): $$1 = e^{-3x}$$ Take the natural logarithm of both sides: $$0 = -3x$$ Now, solve for x: $$x = 0$$ With the intersection point found, we can evaluate the y-value: $$y = e^0 = 1$$ So, the point of intersection is \((0,1)\). Step 2: Sketch the region
02

Plot the given functions and region

First, sketch the curve of \(y = e^x\) (an upward sloping curve) and \(y = e^{-2x}\) (a downward sloping curve). Next, draw a vertical line at \(x = \ln 4\). The region bounded by the three given curves should now be visible. You'll notice that the region is a finite shape enclosed by the three curves. Step 3: Set up the integral representing the area of the region
03

Calculate the integral for the area

To find the area of the region, we will integrate the difference of the two functions along the x-axis from the left intersection point to the rightmost bound, which is \(\ln 4\). Thus, we have: $$A = \int_{0}^{\ln 4} (e^{-2x} - e^x) dx$$ Step 4: Evaluate the integral and find the area
04

Evaluate the integral and find the area

Now, we need to evaluate $$A = \int_{0}^{\ln 4} (e^{-2x} - e^x) dx$$ Integrate the given function with respect to x: $$A = -\frac{1}{2} e^{-2x} - e^x \Big|_{0}^{\ln 4}$$ Plug in the bounds of integration: $$A = -\frac{1}{2} e^{-2(\ln 4)} - e^{\ln 4} -\left(-\frac{1}{2} e^{-2(0)} - e^{0}\right)$$ Simplify the expression: $$A = -\frac{1}{2} \left(\frac{1}{16}\right) - 4 + \frac{1}{2} +1$$ Now, calculate the area: $$A = -\frac{1}{32} - 4 + \frac{1}{2} + 1 = \frac{3}{2} - 4 = -\frac{1}{2}$$ Since the area of the bounded region cannot be negative, we take the absolute value of the result: $$A = |\text{-}\frac{1}{2}| = \frac{1}{2}$$ The area of the region bounded by the curves \(y = e^x\), \(y = e^{-2x}\), and \(x = \ln 4\) is \(\frac{1}{2}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Integration to Find Area
To understand how to find the area bounded by curves, we turn to integration, a fundamental concept in calculus. Integration allows us to accumulate quantities, such as areas, over a certain interval. Specifically, when we want to find the area between two curves, we set up the integral of the difference between the top function and the bottom function over the interval defined by their points of intersection. In our exercise, once we identify the region, we need to integrate the difference between the functions \(y = e^{-2x}\) and \(y = e^x\) from \(x = 0\) to \(x = \ln 4\). This gives us the total area under the upper curve minus the area under the lower curve, thus yielding the area of the uniquely shaped region enclosed by the curves and vertical boundary.
Sketching Regions in Calculus
Visualizing the area between curves with a sketch is a vital step in calculus. Doing so not only helps us conceptualize the problem but also ensures that we integrate over the correct bounds and in the right direction. In this problem, we begin by plotting the two exponential curves \(y = e^{x}\) and \(y = e^{-2x}\), and the vertical line \(x = \ln 4\). We look for the enclosed area and note its boundaries - it's the intersection points that describe where the curves meet, which serve as the limits of integration for our area calculation. These graphical illustrations can be instrumental in avoiding common calculus mistakes like integrating over the wrong interval or misidentifying the functions that bound the region.
Exponential Functions
Exponential functions, represented as \(y = a^x\), where 'a' is a constant and 'x' is the exponent, are instrumental in describing growth or decay processes in calculus. The base of the exponential function in our exercise is 'e', known as Euler's number, which is approximately equal to 2.71828. This special base is crucial because it creates growth rates that are proportional to values at each point, a property used widely in calculus, physics, and finance. In the context of our problem, \(y = e^x\) describes an increasing exponential function, whereas \(y = e^{-2x}\) describes a decreasing one. Their intersection and behavior define the nature of the region we are interested in.
Natural Logarithm Properties
The natural logarithm, denoted as \(\ln\), is the inverse of the exponential function with base 'e'. It comes with a set of properties that simplifies solving equations involving 'e'. One of the most frequently used properties is that \(\ln(e^x) = x\), which we use in the exercise to determine the bounds of integration when one limit is \(\ln 4\). Moreover, the natural logarithm of 1 is always 0, since \(e^0 = 1\), and this is used to determine the intersection point of our exponential functions. Understanding these properties enables us to navigate equations and integrals that involve exponential growth or decay.

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Most popular questions from this chapter

Consider the functions \(f(x)=x^{n}\) and \(g(x)=x^{1 / n},\) where \(n \geq 2\) is a positive integer. a. Graph \(f\) and \(g\) for \(n=2,3,\) and \(4,\) for \(x \geq 0\) b. Give a geometric interpretation of the area function \(A_{n}(x)=\int_{0}^{x}(f(s)-g(s)) d s,\) for \(n=2,3,4, \ldots\) and \(x>0\) c. Find the positive root of \(A_{n}(x)=0\) in terms of \(n\). Does the root increase or decrease with \(n\) ?

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