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Magazine Chapter 5: Problem 65
Find a positive value of \(k\) such that the area under the graph of \(y=e^{2 x}\) over the interval \([0, k]\) is 3 square units.
Short Answer
Expert verified
The positive value of \(k\) is \(\frac{\ln 7}{2}\).
Step by step solution
01
Set up the Integral
The problem requires us to find the area under the curve of the function \( y = e^{2x} \) over the interval \([0, k]\). The area under the curve can be found by integrating the function from \(0\) to \(k\): \[\int_{0}^{k} e^{2x} \, dx\]
02
Integrate the Function
To find the integral of \( e^{2x} \), we use the basic integral rule for exponential functions. The integral of \( e^{ax} \) with respect to \( x \) is \( \frac{1}{a}e^{ax} \) plus a constant of integration. In this case, \( a = 2 \), so the integral is:\[ \int e^{2x} \, dx = \frac{1}{2} e^{2x} + C \]We will evaluate this from \(0\) to \(k\).
03
Evaluate the Definite Integral
Substitute the limits of integration into the indefinite integral:\[ \left. \frac{1}{2}e^{2x} \right|_{0}^{k} = \frac{1}{2}e^{2k} - \frac{1}{2}e^{0} \]Since \(e^{0} = 1\), this simplifies to:\[ \frac{1}{2}e^{2k} - \frac{1}{2} \]
04
Set the Equation Equal to the Given Area
We know that the area is given as 3 square units, so we set up the equation:\[ \frac{1}{2}e^{2k} - \frac{1}{2} = 3 \]
05
Solve for \(k\)
First, add \( \frac{1}{2} \) to both sides:\[ \frac{1}{2}e^{2k} = 3 + \frac{1}{2} \]This simplifies to:\[ \frac{1}{2}e^{2k} = \frac{7}{2} \]Next, multiply both sides by 2 to isolate \(e^{2k}\):\[ e^{2k} = 7 \]Take the natural logarithm of both sides to solve for \(2k\):\[ 2k = \ln(7) \]Finally, divide by 2 to solve for \(k\):\[ k = \frac{\ln 7}{2} \]
06
Verify the Solution
To verify, we can substitute \(k = \frac{\ln 7}{2}\) back into the integral setup and ensure the area equals 3:Calculate the area:\[ \frac{1}{2} e^{2(\frac{\ln 7}{2})} - \frac{1}{2} = \frac{1}{2} \, 7 - \frac{1}{2} = 3\]The calculation confirms that the area is 3 square units.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Definite Integral
The concept of a definite integral is crucial to calculating the area under a curve.
At its core, it involves integrating a function over a specified interval. For example, the integral \[ \int_{a}^{b} f(x) \, dx \] represents the area under the curve of the function \(f(x)\) from \(x = a\) to \(x = b\).
Unlike indefinite integrals, which result in a general formula involving a constant, definite integrals yield specific values.
This process incorporates the limits of integration to specify the start and end points on the x-axis. Once integrated, you can substitute these limits into the antiderivative to find the exact area.
At its core, it involves integrating a function over a specified interval. For example, the integral \[ \int_{a}^{b} f(x) \, dx \] represents the area under the curve of the function \(f(x)\) from \(x = a\) to \(x = b\).
Unlike indefinite integrals, which result in a general formula involving a constant, definite integrals yield specific values.
This process incorporates the limits of integration to specify the start and end points on the x-axis. Once integrated, you can substitute these limits into the antiderivative to find the exact area.
- Set up the integral with the correct limits
- Integrate the function to find the antiderivative
- Apply the limits to evaluate the area
Exponential Functions
Exponential functions, such as \(y = e^{2x}\), are unique due to their constant percentage growth rate in relation to their own value.
This means they grow or decay at a rate proportional to their current size, leading to exponential growth or decay.
The integral of an exponential function often follows a standard formula. When integrating a function of the form \(e^{ax}\), the antiderivative is \(\frac{1}{a}e^{ax} + C\), where \(C\) is the constant of integration.
This means they grow or decay at a rate proportional to their current size, leading to exponential growth or decay.
The integral of an exponential function often follows a standard formula. When integrating a function of the form \(e^{ax}\), the antiderivative is \(\frac{1}{a}e^{ax} + C\), where \(C\) is the constant of integration.
- The base "e" is approximately 2.718 and is a mathematical constant
- "a" is the coefficient that influences the growth rate in \(e^{ax}\)
Area Under a Curve
Finding the area under a curve is a primary application of integration.
It involves determining the total accumulation or net change represented by the curve, bounded by specific points on the x-axis. When you want to find the area under the curve for functions, the definite integral is the tool to use.
For instance, when you compute the integral of \(y = e^{2x}\) from 0 to \(k\), you're calculating how much area that curve encloses above the x-axis to the left of \(k\).
It involves determining the total accumulation or net change represented by the curve, bounded by specific points on the x-axis. When you want to find the area under the curve for functions, the definite integral is the tool to use.
For instance, when you compute the integral of \(y = e^{2x}\) from 0 to \(k\), you're calculating how much area that curve encloses above the x-axis to the left of \(k\).
- Helps in real-life applications like calculating distance traveled, revenue over time, or material needed
- Considers positive contributions of area in the desired interval
