Chapter 12: Problem 58
Find the area of the region bounded by the graphs of the given equations. $$ \begin{aligned} &y=\frac{1}{9} x e^{-x / 3}\\\ &y=0, x=0, x=3 \end{aligned} $$
Short Answer
Expert verified
The area of the bounded region is \( \frac{e ^ {-1}}{3} + 1\) square units.
Step by step solution
01
Identify the equations
The equations give the boundaries of the region we are finding the area of. y=0 is the x-axis, \(x=0\) is the y-axis, and \(x=3\) is the vertical line at \(x=3\). The function \(y=\frac{1}{9} x e^{-x / 3}\) is a decreasing exponential function starting at the origin (0,0).
02
Set up the definite integral
The area of this bounded region can be calculated as the definite integral from \(x=0\) to \(x=3\) (the limits given by \(x\)) of the function \(y(x) = \frac{1}{9} x e^{-x / 3}\) that is, \[ \int_0 ^3 \frac{1}{9} x e^{-x / 3} dx \]
03
Solve the definite integral
To compute this area, we solve the integral by using Integration by parts, where \(u = x\) and \(dv = \frac{1}{9}e^{-x / 3}dx\). Following the formula for integration by parts, we get \[ \int u dv = uv - \int v du \] After computing, the integral becomes \( -x e^{-x / 3} - e^{-x / 3} \]
04
Evaluate and simplify
Evaluate the computed integral between \(x=0\) to \(x=3\). After plugging in the limits and subtracting, we obtain The final simplified equation is \( \frac{e ^ {-1}}{3} + 1 \)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integration by Parts
Integration by parts is a fundamental technique in calculus used to integrate products of two functions. It is derived from the product rule of differentiation and provides a way to break down more complex integrals into simpler forms.
The formula for integration by parts is given by: \[ \text{If } u = f(x) \text{ and } dv = g(x)dx, \text{ then } \frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}, \] where after integrating both sides with respect to \(x\), we get \[ \text{Integration by Parts: } \text{Standard Form: } \text{If } u = f(x) \text{ and } dv = g(x)dx, \text{ then } \int u dv = uv - \int v du. \]To apply this method, we choose \(u\) and \(dv\) such that the resulting integral \(\int v du\) is easier to solve than the original.
In the given exercise, by choosing \(u = x\) and \(dv = \frac{1}{9}e^{-x / 3}dx\), the integral was simplified to a form that could be easily evaluated. This technique not only helps in solving the integral at hand but also strengthens our understanding of the interplay between differentiation and integration.
The formula for integration by parts is given by: \[ \text{If } u = f(x) \text{ and } dv = g(x)dx, \text{ then } \frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}, \] where after integrating both sides with respect to \(x\), we get \[ \text{Integration by Parts: } \text{Standard Form: } \text{If } u = f(x) \text{ and } dv = g(x)dx, \text{ then } \int u dv = uv - \int v du. \]To apply this method, we choose \(u\) and \(dv\) such that the resulting integral \(\int v du\) is easier to solve than the original.
In the given exercise, by choosing \(u = x\) and \(dv = \frac{1}{9}e^{-x / 3}dx\), the integral was simplified to a form that could be easily evaluated. This technique not only helps in solving the integral at hand but also strengthens our understanding of the interplay between differentiation and integration.
Exponential Functions
Exponential functions are mathematical functions of the form \(f(x) = a^{bx}\), where \(a\) is a positive constant base, and \(b\) is the exponent as a function of \(x\). In calculus, we often deal with the special case where \(a = e\), where \(e\) represents Euler's number, approximately equal to 2.71828.
These functions exhibit rapid growth or decay, which is mirrored in real-world phenomena like population growth, radioactive decay, and compound interest. They have unique properties: a constant rate of growth or decay (percentage change) and a graph that is always positive. For the function in our exercise, \(y = \frac{1}{9} x e^{-x / 3}\), the exponential term \(e^{-x / 3}\) signifies a decay factor, and it decreases as \(x\) increases. The presence of \(x\) in the numerator introduces a variance in the rate of growth that distinguishes it from the standard exponential form.
The behavior of exponential functions is essential to calculus, especially when we are dealing with changes that follow a continuous rate. In finding areas under curves, understanding the rate at which the function decreases helps us predict the shape of the area we are calculating.
These functions exhibit rapid growth or decay, which is mirrored in real-world phenomena like population growth, radioactive decay, and compound interest. They have unique properties: a constant rate of growth or decay (percentage change) and a graph that is always positive. For the function in our exercise, \(y = \frac{1}{9} x e^{-x / 3}\), the exponential term \(e^{-x / 3}\) signifies a decay factor, and it decreases as \(x\) increases. The presence of \(x\) in the numerator introduces a variance in the rate of growth that distinguishes it from the standard exponential form.
The behavior of exponential functions is essential to calculus, especially when we are dealing with changes that follow a continuous rate. In finding areas under curves, understanding the rate at which the function decreases helps us predict the shape of the area we are calculating.
Area Under a Curve
The concept of the area under a curve in calculus is a fundamental application of the definite integral. The area represents the accumulation of quantities over a certain interval, and it's one of the most visual representations of integral calculus.
Formally, if \(f(x)\) is a continuous and non-negative function on the interval \([a, b]\), the area under the curve from \(a\) to \(b\) is given by the definite integral: \[ A = \int_{a}^{b} f(x) \, dx.\]
A definite integral has both upper and lower limits, and in the context of the given exercise, the limits are from \(x = 0\) to \(x = 3\). When we evaluate this definite integral, we get the exact area enclosed between the function's graph and the \(x\)-axis over the specified interval.
The point of intersection with the axes and the behavior of the function within the defined limits are pivotal in determining the area. The region's boundaries given by the equations in the exercise \(y=0\), \(x=0\), and \(x=3\) with the function \(y=\frac{1}{9} x e^{-x / 3}\) define the area that we need to calculate. Understanding the shape and the limiting behavior of the function helps to comprehend the integral's result and ensures that we grasp not just the 'how' but also the 'why' behind the process of finding the area under a curve.
Formally, if \(f(x)\) is a continuous and non-negative function on the interval \([a, b]\), the area under the curve from \(a\) to \(b\) is given by the definite integral: \[ A = \int_{a}^{b} f(x) \, dx.\]
A definite integral has both upper and lower limits, and in the context of the given exercise, the limits are from \(x = 0\) to \(x = 3\). When we evaluate this definite integral, we get the exact area enclosed between the function's graph and the \(x\)-axis over the specified interval.
The point of intersection with the axes and the behavior of the function within the defined limits are pivotal in determining the area. The region's boundaries given by the equations in the exercise \(y=0\), \(x=0\), and \(x=3\) with the function \(y=\frac{1}{9} x e^{-x / 3}\) define the area that we need to calculate. Understanding the shape and the limiting behavior of the function helps to comprehend the integral's result and ensures that we grasp not just the 'how' but also the 'why' behind the process of finding the area under a curve.
