/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 35 Find the area of the indicated r... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the area of the indicated region. We suggest you graph the curves to check whether one is above the other or whether they cross, and that you use technology to check your answers. Enclosed by \(y=e^{x}, y=2\), and the \(y\) -axis

Short Answer

Expert verified
The area of the region enclosed by the curves \(y = e^x\), \(y = 2\), and the y-axis is \(2\ln(2) - 1\) square units.

Step by step solution

01

Determine intersection points between the curves and the y-axis

To find the intersection points between the curves \(y = e^x\), \(y = 2\), and the y-axis, we need to solve the following equations: 1. Intersection between \(y = e^x\) and \(y = 2\): We need to find the value of x that satisfies both equations. Therefore, we have to solve the equation: \(e^x = 2\) 2. Intersection between \(y = e^x\) and the y-axis: The y-axis is at x = 0. Therefore, we have to find the value of y that satisfies the equation: \(y = e^0\) 3. Intersection between \(y = 2\) and the y-axis: The y-axis is at x = 0. Therefore, we have to find the value of y that satisfies the equation: \(y = 2\)
02

Solve equations to find the intersection points

Now we will solve the equations defined in Step 1: 1. Intersection between \(y = e^x\) and \(y = 2\): \(e^x = 2\) To solve for x, we take the natural logarithm of both sides: \(x = \ln(2)\) Therefore, the intersection point between the curves is at: \(x = \ln(2)\) 2. Intersection between \(y = e^x\) and the y-axis: \(y = e^0\) Since anything raised to the power of 0 is 1: \(y = 1\) Therefore, the intersection point between the curve \(y = e^x\) and the y-axis is at: \(y = 1\) 3. Intersection between \(y = 2\) and the y-axis: The curve \(y = 2\) intersects the y-axis at: \(y = 2\)
03

Set up the integral to find the area

We will now set up the integral to find the area of the region enclosed by the curves. To do this, we will subtract the lower curve from the upper curve and integrate with respect to x. Since the curve \(y = 2\) is above the curve \(y = e^x\) in the region of interest, we will subtract the equation of the lower curve (y = e^x) from the equation of the upper curve (y = 2). The limits of integration will be from x = 0 (the y-axis) to x = ln(2) (the intersection point between the curves): Area = \(\int_{0}^{\ln(2)} (2 - e^x) dx\)
04

Evaluate the integral

Now we will evaluate the integral to find the area of the region enclosed by the curves: Area = \(\int_{0}^{\ln(2)} (2 - e^x) dx\) = \([2x - e^x]_{0}^{\ln(2)}\) = \((2\ln(2) - e^{\ln(2)}) - (2(0) - e^0)\) = \((2\ln(2) - 2) - (0 - 1)\) = \(2\ln(2) - 1\) The area of the region enclosed by the curves \(y = e^x\), \(y = 2\), and the y-axis is \(2\ln(2) - 1\) square units.

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Key Concepts

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

Integration
Integration is a crucial concept in calculus that helps us find areas under curves and accumulate quantities like distance or mass. In essence, integration is the reverse process of differentiation. Just like how differentiation gives us the rate of change, integration accumulates these changes over an interval.

When solving problems involving the area under curves, integration allows us to sum infinitesimally small slices of area to find the total area over a specific interval. This is particularly useful when the boundary is not a simple shape.
  • The integral is expressed as \([\int f(x) \, dx]\), representing the area under the curve of \(f(x)\) with respect to \(x\).
  • The limits of integration define the interval over which you're computing the area, from start (lower limit) to end (upper limit).
Understanding integration is essential for calculating more complex shapes and boundary conditions in physics, engineering, and beyond.
Area under curve
Finding the area under a curve is a common application of integration. This technique is used to calculate the total area that lies between a curve and the horizontal axis (or another reference line) over a specific interval.

In the context of the provided exercise, we are searching for the area enclosed by the exponential function \(y = e^x\), the constant line \(y = 2\), and the \(y\)-axis.
For this:
  • You first determine which curve is the upper curve and which is the lower, based on their positions relative to each other over the chosen interval.
  • The area is then computed by integrating the difference between the upper and lower curves over the interval where they enclose the region.
Think of the process as cutting out the shape from the space under the upper curve down to the lower one, giving you the exact enclosed area.
Exponential functions
Exponential functions are functions where the variable is an exponent. They commonly appear in the form \(y = a\cdot b^{x}\), where \(b > 0\) and \(b eq 1\). In many real-world applications, we see these functions modeling growth processes like population or decay processes like radioactive decay.

In this exercise, the specific exponential function is \(y = e^x\), which uses the base \(e\), the natural exponential constant approximately equal to 2.71828. This base is special because it simplifies the calculus associated with growth and decay.
  • Exponential growth: As \(x\) increases, \(e^x\) increases rapidly.
  • Exponential decay: If a negative exponent is used, such as \(e^{-x}\), the function decreases quickly as \(x\) increases.
Understanding these functions is vital as they provide models for many natural phenomena and are involved in various scientific and financial calculations.

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