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Magazine Chapter 5: Problem 13
Find the area of the regions.Under \(y=e^{x}\) and above \(y=1\) for \(0 \leq x \leq 2\).
Short Answer
Expert verified
The area is \( e^2 - 3 \).
Step by step solution
01
Identify the Area to Calculate
The area we need to find is the region under the curve of the function \( y = e^x \) and above the line \( y = 1 \) within the interval \( 0 \leq x \leq 2 \).
02
Set Up the Integral Expression
To find the area between the curves, set up the integral expression. The function \( y = e^x \) is greater than \( y = 1 \) in this interval, so the area can be expressed as: \[\text{Area} = \int_0^2 (e^x - 1) \, dx\]
03
Integrate the Expression
Compute the definite integral. First, integrate \( e^x \) and \( 1 \) separately:\[ \int e^x \, dx = e^x \]\[ \int 1 \, dx = x \] Thus, the integral becomes:\[ \int_0^2 (e^x - 1) \, dx = \left[ e^x - x \right]_0^2\]
04
Evaluate the Integral
Now substitute the boundaries into the integrated expression:\[ \left[ e^x - x \right]_0^2 = \left( e^2 - 2 \right) - \left( e^0 - 0 \right) \]\[ = (e^2 - 2) - (1 - 0)\]\[= e^2 - 2 - 1\]\[ = e^2 - 3\]
05
Final Calculation
The final result of the integral gives the area of the region:\[ \text{Area} = e^2 - 3 \] where \( e \) is the base of the natural logarithm, approximately 2.71828.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding the Definite Integral
The definite integral helps us calculate the area under a curve within a specified interval on the x-axis. In this exercise, we calculate this area between two curves, which are given by the function \( y = e^x \) and a constant \( y = 1 \).
To understand how this works, imagine plotting the function \( y = e^x \) and the line \( y = 1 \) on a graph. The region of interest is the space between these two within the limits \( 0 \) and \( 2 \). Using integrals, we can calculate this space by integrating the "gap" between \( e^x \) and \( 1 \), effectively slicing the area into infinitely small slivers and summing them up using the integral:
To understand how this works, imagine plotting the function \( y = e^x \) and the line \( y = 1 \) on a graph. The region of interest is the space between these two within the limits \( 0 \) and \( 2 \). Using integrals, we can calculate this space by integrating the "gap" between \( e^x \) and \( 1 \), effectively slicing the area into infinitely small slivers and summing them up using the integral:
- The integral \( \int_0^2 e^x \) calculates the total area under the curve \( y = e^x \) from \( 0 \) to \( 2 \).
- The integral \( \int_0^2 1 \) represents the area of the rectangle formed by the constant line \( y = 1 \) over the same interval.
- Subtracting these gives the desired area, \( \int_0^2 (e^x - 1) \, dx \).
Diving into Exponential Functions
Exponential functions are functions of the form \( y = a^x \), where \( a \) is a constant. In our exercise, the function is specifically \( y = e^x \) with \( e \) being an irrational constant approximately equal to 2.71828.
- Exponential functions exhibit continuous and rapid growth, making them different from linear functions.
- The function \( y = e^x \) increases rapidly as \( x \) increases, creating a curve whose slope becomes steeper and steeper.
- These functions model real-world phenomena that grow at rates proportional to their size, like population growth and compound interest. Hence, understanding them is vital!
Exploring Integration Techniques
Integration is a powerful tool for finding areas under curves. To tackle this problem, we employ basic integration techniques. Here, we break it down:
Evaluating this definite integral involves substituting the boundary values, giving a concrete number that provides the precise area under the curve between the specified lines or curves over the interval in question!
- First, integrate the exponential function \( e^x \), which is rather straightforward since its antiderivative is simply \( e^x \) itself.
- Next, integrate the constant function \( 1 \). The antiderivative of a constant is the constant multiplied by the variable, which results in \( x \).
Evaluating this definite integral involves substituting the boundary values, giving a concrete number that provides the precise area under the curve between the specified lines or curves over the interval in question!
