Let The Region \[$ R \$\] Be The Area Enclosed By The Function \($f(x) = E^x + 1$\), The Horizontal Line \($y = -1$\), And The Vertical Lines \[$x = 0$\$\] And \($x = 2$\).Find The Volume Of The

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Let the Region { R $}$ be the Area Enclosed by the Function {f(x) = e^x + 1}, the Horizontal Line {y = -1}, and the Vertical Lines {x=0x = 0$}$ and {x = 2}. Find the Volume of the Region { R $}$

In this article, we will explore the concept of finding the volume of a region enclosed by a function, a horizontal line, and two vertical lines. The function in question is {f(x) = e^x + 1}, the horizontal line is {y = -1}, and the vertical lines are {x=0x = 0$}$ and {x = 2}. We will use the method of integration to find the volume of the region enclosed by these boundaries.

Understanding the Region { R $}$

To begin, let's visualize the region { R $}$ that we are interested in finding the volume of. The function {f(x) = e^x + 1} is an exponential function that increases as {x}$ increases. The horizontal line {y = -1} is a constant line that intersects the function at a single point. The vertical lines {x=0x = 0$}$ and (x=2x = 2] are the boundaries of the region in the x-direction.

Finding the Volume of the Region { R $}$

To find the volume of the region { R $}$, we will use the method of integration. We will integrate the area under the curve of the function {f(x) = e^x + 1} with respect to {x}$ from {x = 0} to {x = 2}. This will give us the volume of the region enclosed by the function, the horizontal line, and the vertical lines.

Step 1: Define the Function and the Limits of Integration

The function that we are interested in integrating is {f(x) = e^x + 1}. The limits of integration are {x = 0} and {x = 2}.

Step 2: Integrate the Function

To find the volume of the region, we will integrate the function {f(x) = e^x + 1} with respect to {x}$ from {x = 0} to {x = 2}. This can be done using the following integral:

∫02(ex+1)dx\int_{0}^{2} (e^x + 1) dx

Step 3: Evaluate the Integral

To evaluate the integral, we will use the following formula:

∫(ex+1)dx=ex+x+C\int (e^x + 1) dx = e^x + x + C

where {C}$ is the constant of integration.

Step 4: Apply the Limits of Integration

To find the volume of the region, we will apply the limits of integration to the integral:

∫02(ex+1)dx=[ex+x]02\int_{0}^{2} (e^x + 1) dx = \left[e^x + x\right]_{0}^{2}

Step 5: Simplify the Expression

To simplify the expression, we will evaluate the expression at the limits of integration:

[ex+x]02=(e2+2)−(e0+0)\left[e^x + x\right]_{0}^{2} = (e^2 + 2) - (e^0 + 0)

Step 6: Simplify Further

To simplify further, we will use the following properties of exponents:

e0=1e^0 = 1

e2=e2−0=e2e^2 = e^{2-0} = e^2

Step 7: Simplify the Expression

To simplify the expression, we will substitute the values of {e^0}$ and {e^2}$ into the expression:

(e2+2)−(e0+0)=(e2+2)−1(e^2 + 2) - (e^0 + 0) = (e^2 + 2) - 1

Step 8: Simplify Further

To simplify further, we will combine like terms:

(e2+2)−1=e2+1(e^2 + 2) - 1 = e^2 + 1

In this article, we have found the volume of the region enclosed by the function {f(x) = e^x + 1}, the horizontal line {y = -1}, and the vertical lines {x=0x = 0$}$ and {x = 2}. The volume of the region is given by the integral:

∫02(ex+1)dx=e2+1\int_{0}^{2} (e^x + 1) dx = e^2 + 1

This result can be used to find the volume of similar regions enclosed by exponential functions and horizontal lines.

  • [1] Calculus, 3rd edition, Michael Spivak
  • [2] Calculus, 2nd edition, James Stewart
  • [3] Calculus, 1st edition, Michael Spivak
  • Exponential function: A function of the form {f(x) = a^x}$, where {a}$ is a positive constant.
  • Horizontal line: A line of the form {y = c}$, where {c}$ is a constant.
  • Vertical line: A line of the form {x = c}$, where {c}$ is a constant.
  • Integral: A mathematical operation that finds the area under a curve.
  • Limits of integration: The values of {x}$ at which the integral is evaluated.
  • Constant of integration: A constant that is added to the integral to make it equal to the original function.
    Q&A: Finding the Volume of the Region Enclosed by the Function {f(x) = e^x + 1}, the Horizontal Line {y = -1}, and the Vertical Lines {x=0x = 0$}$ and {x = 2}

Q: What is the region { R $}$ that we are interested in finding the volume of?

A: The region { R $}$ is the area enclosed by the function {f(x) = e^x + 1}, the horizontal line {y = -1}, and the vertical lines {x=0x = 0$}$ and {x = 2}.

Q: Why do we need to find the volume of the region { R $}$?

A: We need to find the volume of the region { R $}$ because it represents the amount of space enclosed by the function, the horizontal line, and the vertical lines.

Q: What is the method of integration that we will use to find the volume of the region { R $}$?

A: We will use the method of integration to find the volume of the region { R $}$. This involves integrating the area under the curve of the function {f(x) = e^x + 1} with respect to {x}$ from {x = 0} to {x = 2}.

Q: What is the integral that we will evaluate to find the volume of the region { R $}$?

A: The integral that we will evaluate to find the volume of the region { R $}$ is:

∫02(ex+1)dx\int_{0}^{2} (e^x + 1) dx

Q: How do we evaluate the integral?

A: To evaluate the integral, we will use the following formula:

∫(ex+1)dx=ex+x+C\int (e^x + 1) dx = e^x + x + C

where {C}$ is the constant of integration.

Q: What are the limits of integration that we will apply to the integral?

A: The limits of integration that we will apply to the integral are {x = 0} and {x = 2}.

Q: How do we simplify the expression after applying the limits of integration?

A: To simplify the expression, we will evaluate the expression at the limits of integration:

[ex+x]02=(e2+2)−(e0+0)\left[e^x + x\right]_{0}^{2} = (e^2 + 2) - (e^0 + 0)

Q: What is the final answer for the volume of the region { R $}$?

A: The final answer for the volume of the region { R $}$ is:

e2+1e^2 + 1

Q: What is the significance of the result?

A: The result represents the volume of the region enclosed by the function {f(x) = e^x + 1}, the horizontal line {y = -1}, and the vertical lines {x=0x = 0$}$ and {x = 2}.

Q: Can we use this result to find the volume of similar regions?

A: Yes, we can use this result to find the volume of similar regions enclosed by exponential functions and horizontal lines.

Q: What are some common applications of finding the volume of regions?

A: Some common applications of finding the volume of regions include:

  • Calculating the volume of objects in engineering and architecture
  • Finding the volume of fluids in physics and chemistry
  • Calculating the volume of gases in thermodynamics

Q: What are some common challenges in finding the volume of regions?

A: Some common challenges in finding the volume of regions include:

  • Choosing the correct method of integration
  • Evaluating the integral correctly
  • Applying the limits of integration correctly

Q: How can we overcome these challenges?

A: We can overcome these challenges by:

  • Choosing the correct method of integration
  • Evaluating the integral carefully
  • Applying the limits of integration correctly

In this Q&A article, we have discussed the method of integration that we will use to find the volume of the region enclosed by the function {f(x) = e^x + 1}, the horizontal line {y = -1}, and the vertical lines {x=0x = 0$}$ and {x = 2}. We have also discussed the integral that we will evaluate to find the volume of the region, and the final answer for the volume of the region. We have also discussed some common applications and challenges in finding the volume of regions.