Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 9: Problem 49
Find the centroid of the region bounded by the graphs of the equations \(y=e^{x}, y=0, x=0,\) and \(x=\ln 3\).
Short Answer
Expert verified
The centroid is \(\left(\frac{3 \ln 3 - 2}{2}, 1\right).\)
Step by step solution
01
Understanding the Region
First, identify the region bounded by the graphs. The region is bounded by the curve \(y = e^{x}\), the x-axis \(y = 0\), the line \(x = 0\), and the line \(x = \ln 3\). This is a vertical strip from \(x=0\) to \(x=\ln 3\) under the curve \(y=e^x\).
02
Formula for the Centroid
The formula for the centroid \((\bar{x}, \bar{y})\) of a region under a curve \(y=f(x)\) from \(x=a\) to \(x=b\) is given by:\[ \bar{x} = \frac{1}{A} \int_{a}^{b} x f(x) \, dx \] \[ \bar{y} = \frac{1}{A} \int_{a}^{b} \frac{1}{2} [f(x)]^2 \, dx \]where \(A\) is the area under the curve from \(x=a\) to \(x=b\).
03
Calculating the Area of the Region \(A\)
The area \(A\) under \(y=e^x\) from \(x=0\) to \(x=\ln 3\) is:\[ A = \int_{0}^{\ln 3} e^x \, dx = [e^x]_{0}^{\ln 3} = e^{\ln 3} - e^0 = 3 - 1 = 2. \]
04
Calculating \(\bar{x}\)
Now calculate \(\bar{x}\):\[ \bar{x} = \frac{1}{2} \int_{0}^{\ln 3} x e^x \, dx. \]Use integration by parts to solve \(\int x e^x \, dx\). Let \(u=x\) and \(dv=e^x \, dx\), then \(du=dx\) and \(v=e^x\). Using integration by parts:\[ \int x e^x \, dx = x e^x - \int e^x \, dx = x e^x - e^x. \]Evaluate from 0 to \(\ln 3\):\[ [x e^x - e^x]_{0}^{\ln 3} = (\ln 3 \cdot 3 - 3) - (0 - 1) = 3 \ln 3 - 2. \]Then,\[ \bar{x} = \frac{1}{2} (3 \ln 3 - 2) = \frac{3 \ln 3 - 2}{2}. \]
05
Calculating \(\bar{y}\)
Now calculate \(\bar{y}\):\[ \bar{y} = \frac{1}{2 \, A} \int_{0}^{\ln 3} [e^x]^2 \, dx = \frac{1}{4} \int_{0}^{\ln 3} e^{2x} \, dx. \]Solve the integral:\[ \int e^{2x} \, dx = \frac{1}{2} e^{2x}. \]Evaluate from 0 to \(\ln 3\):\[ \left[ \frac{1}{2} e^{2x} \right]_{0}^{\ln 3} = \frac{1}{2} (e^{2 \ln 3} - 1) = \frac{1}{2} (3^2 - 1) = \frac{1}{2} \cdot 8 = 4. \]Thus,\[ \bar{y} = \frac{1}{4} \cdot 4 = 1. \]
06
Conclusion
Thus, the centroid of the region is \(\left(\frac{3 \ln 3 - 2}{2}, 1\right).\)
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91影视!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integration by Parts
Integration by parts is a technique used to solve integrals, especially when dealing with products of functions. Think of it as the integrative equivalent of the product rule for differentiation. If you have a function that is difficult to integrate directly, this method can help. The technique is based on the formula:\[ \int u \, dv = uv - \int v \, du \]Here鈥檚 how it works:
- Select two parts from your integrand: one to differentiate (\(u\)) and the other to integrate (\(dv\)).
- Calculate the derivative of \(u\) to get \(du\), and integrate \(dv\) to obtain \(v\).
- Substitute these into the formula to get the integral that might be easier to evaluate.
Area Under a Curve
Finding the area under a curve is fundamental for many applications in mathematics. It鈥檚 essentially the integral of the function that describes the curve over a specified interval. To find this area, we evaluate the definite integral of the function.In this exercise, we're dealing with the exponential function \(y = e^x\), bounded by vertical lines \(x = 0\) and \(x = \ln 3\), and the x-axis. To find the area, we calculate:\[ A = \int_{0}^{\ln 3} e^x \, dx \]Here鈥檚 a breakdown:
- Integrate \(e^x\) over the interval from 0 to \(\ln 3\).
- The result is \([e^x]_{0}^{\ln 3} = e^{\ln 3} - e^0 = 3 - 1\).
- This simplifies to 2, depicting the area under the curve over the given interval.
Exponential Functions
Exponential functions, like the function \(y = e^x\) used in this problem, are very significant in many areas of math and science. These functions have the form \(f(x) = a^x\), where \(a\) is a constant, \(a > 0\), and \(a eq 1\). The function \(e^x\), in particular, is special because \(e\), approximately 2.718, is the base of natural logarithms.Why are exponential functions important?
- They model growth and decay in natural processes like population increase or radioactive decay.
- They feature prominently in financial mathematics, predicting compound interest.
- The slope of \(e^x\) is equal to its value at any point \(x\), meaning its rate of change is proportional to its current value.
Definite Integrals
Definite integrals are a core concept in calculus, helping us calculate the exact area under a curve from one point to another on the x-axis. When you see \(\int_{a}^{b} f(x) \, dx\), it tells you to find the area under the curve \(f(x)\) from \(x = a\) to \(x = b\).Essential steps for solving definite integrals:
- Find the antiderivative of the function \(f(x)\), denoted as \(F(x)\).
- Evaluate \(F(x)\) at both \(x = a\) and \(x = b\).
- Subtract \(F(a)\) from \(F(b)\), i.e., \(F(b) - F(a)\).
