Evaluate The Volume V V V Using The Given Integral:${ V = \pi \int_0^{-\ln \frac{1}{2}} \left[ \left( 1 - E^{-x} \right)^2 - \left( E^{-x} - 4.5 \right)^2 \right] , Dx }$

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Introduction

In mathematics, the volume of a solid can be calculated using a variety of methods, including the use of definite integrals. In this article, we will evaluate the volume of a solid using the given integral:

V=π∫0−ln⁡12[(1−e−x)2−(e−x−4.5)2] dx{ V = \pi \int_0^{-\ln \frac{1}{2}} \left[ \left( 1 - e^{-x} \right)^2 - \left( e^{-x} - 4.5 \right)^2 \right] \, dx }

Understanding the Integral

The given integral represents the volume of a solid, and it is defined as the product of the constant π and the definite integral of a function. The function inside the integral is a difference of two squared expressions, which can be expanded and simplified.

Expanding and Simplifying the Integral

To evaluate the integral, we first need to expand and simplify the expression inside the integral. We can do this by using the formula for the difference of squares:

(a−b)2−(c−d)2=(a−c)(a+c)−(b−d)(b+d){ (a - b)^2 - (c - d)^2 = (a - c)(a + c) - (b - d)(b + d) }

Applying this formula to the given integral, we get:

(1−e−x)2−(e−x−4.5)2=(1−e−x−e−x+4.5)(1−e−x+e−x+4.5)−(e−x−4.5+e−x+4.5)(e−x−4.5−e−x+4.5){ \left( 1 - e^{-x} \right)^2 - \left( e^{-x} - 4.5 \right)^2 = (1 - e^{-x} - e^{-x} + 4.5)(1 - e^{-x} + e^{-x} + 4.5) - (e^{-x} - 4.5 + e^{-x} + 4.5)(e^{-x} - 4.5 - e^{-x} + 4.5) }

Simplifying this expression, we get:

(1−2e−x+4.5)(1+2e−x+4.5)−(2e−x+8.5)(−2e−x+8.5){ (1 - 2e^{-x} + 4.5)(1 + 2e^{-x} + 4.5) - (2e^{-x} + 8.5)(-2e^{-x} + 8.5) }

Evaluating the Integral

Now that we have simplified the expression inside the integral, we can evaluate the integral. We can do this by using the following properties of definite integrals:

  • The integral of a constant is the constant times the upper limit of integration minus the constant times the lower limit of integration.
  • The integral of a product of two functions is the product of the integrals of the two functions.

Using these properties, we can evaluate the integral as follows:

V=π∫0−ln⁡12[(1−2e−x+4.5)(1+2e−x+4.5)−(2e−x+8.5)(−2e−x+8.5)] dx{ V = \pi \int_0^{-\ln \frac{1}{2}} \left[ (1 - 2e^{-x} + 4.5)(1 + 2e^{-x} + 4.5) - (2e^{-x} + 8.5)(-2e^{-x} + 8.5) \right] \, dx }

V=π[∫0−ln⁡12(1−2e−x+4.5)(1+2e−x+4.5) dx−∫0−ln⁡12(2e−x+8.5)(−2e−x+8.5) dx]{ V = \pi \left[ \int_0^{-\ln \frac{1}{2}} (1 - 2e^{-x} + 4.5)(1 + 2e^{-x} + 4.5) \, dx - \int_0^{-\ln \frac{1}{2}} (2e^{-x} + 8.5)(-2e^{-x} + 8.5) \, dx \right] }

Solving the First Integral

To solve the first integral, we can use the following property of definite integrals:

  • The integral of a product of two functions is the product of the integrals of the two functions.

Using this property, we can rewrite the first integral as:

∫0−ln⁡12(1−2e−x+4.5)(1+2e−x+4.5) dx=∫0−ln⁡12(1+4.5)2−(2e−x)2 dx{ \int_0^{-\ln \frac{1}{2}} (1 - 2e^{-x} + 4.5)(1 + 2e^{-x} + 4.5) \, dx = \int_0^{-\ln \frac{1}{2}} (1 + 4.5)^2 - (2e^{-x})^2 \, dx }

=∫0−ln⁡1220.25−4e−2x dx{ = \int_0^{-\ln \frac{1}{2}} 20.25 - 4e^{-2x} \, dx }

Solving the Second Integral

To solve the second integral, we can use the following property of definite integrals:

  • The integral of a product of two functions is the product of the integrals of the two functions.

Using this property, we can rewrite the second integral as:

∫0−ln⁡12(2e−x+8.5)(−2e−x+8.5) dx=∫0−ln⁡12(−4e−2x+72.25) dx{ \int_0^{-\ln \frac{1}{2}} (2e^{-x} + 8.5)(-2e^{-x} + 8.5) \, dx = \int_0^{-\ln \frac{1}{2}} (-4e^{-2x} + 72.25) \, dx }

Evaluating the Integrals

Now that we have rewritten the integrals, we can evaluate them. We can do this by using the following properties of definite integrals:

  • The integral of a constant is the constant times the upper limit of integration minus the constant times the lower limit of integration.
  • The integral of an exponential function is the exponential function evaluated at the upper limit of integration minus the exponential function evaluated at the lower limit of integration.

Using these properties, we can evaluate the integrals as follows:

∫0−ln⁡1220.25−4e−2x dx=20.25x+2e−2x∣0−ln⁡12{ \int_0^{-\ln \frac{1}{2}} 20.25 - 4e^{-2x} \, dx = 20.25x + 2e^{-2x} \Big|_0^{-\ln \frac{1}{2}} }

=20.25(−ln⁡12)+2e2ln⁡12−(20.25(0)+2e2(0)){ = 20.25(-\ln \frac{1}{2}) + 2e^{2\ln \frac{1}{2}} - (20.25(0) + 2e^{2(0)}) }

=−20.25ln⁡12+2(12)2−2{ = -20.25\ln \frac{1}{2} + 2(\frac{1}{2})^2 - 2 }

=−20.25ln⁡12+12−2{ = -20.25\ln \frac{1}{2} + \frac{1}{2} - 2 }

=−20.25ln⁡12−32{ = -20.25\ln \frac{1}{2} - \frac{3}{2} }

∫0−ln⁡12(−4e−2x+72.25) dx=−2e−2x+72.25x∣0−ln⁡12{ \int_0^{-\ln \frac{1}{2}} (-4e^{-2x} + 72.25) \, dx = -2e^{-2x} + 72.25x \Big|_0^{-\ln \frac{1}{2}} }

=−2e2ln⁡12+72.25(−ln⁡12)−(−2e2(0)+72.25(0)){ = -2e^{2\ln \frac{1}{2}} + 72.25(-\ln \frac{1}{2}) - (-2e^{2(0)} + 72.25(0)) }

=−2(12)2+72.25(−ln⁡12)+2−0{ = -2(\frac{1}{2})^2 + 72.25(-\ln \frac{1}{2}) + 2 - 0 }

=−12+72.25(−ln⁡12)+2{ = -\frac{1}{2} + 72.25(-\ln \frac{1}{2}) + 2 }

=72.25(−ln⁡12)+32{ = 72.25(-\ln \frac{1}{2}) + \frac{3}{2} }

Substituting the Results

Now that we have evaluated the integrals, we can substitute the results into the original equation:

V=π[−20.25ln⁡12−32−72.25(−ln⁡12)−32]{ V = \pi \left[ -20.25\ln \frac{1}{2} - \frac{3}{2} - 72.25(-\ln \frac{1}{2}) - \frac{3}{2} \right] }

V=π[−20.25ln⁡12+72.25(−ln⁡12)−3]{ V = \pi \left[ -20.25\ln \frac{1}{2} + 72.25(-\ln \frac{1}{2}) - 3 \right] }

V=π[−92.5ln⁡12−3]{ V = \pi \left[ -92.5\ln \frac{1}{2} - 3 \right] }

Simplifying the Result

Now that we have substituted the results into the original equation, we can simplify the result:

V=π[−92.5ln⁡12−3]{ V = \pi \left[ -92.5\ln \frac{1}{2} - 3 \right] }

V=−92.5πln⁡12−3π{ V = -92.5\pi \ln \frac{1}{2} - 3\pi }

V=−92.5πln⁡12−3π{ V = -92.5\pi \ln \frac{1}{2} - 3\pi }

Conclusion

In this article, we evaluated the volume of a solid using the given integral:

V=π∫0−ln⁡12[(1−e−x)2−(e−x−4.5)2] dx{ V = \pi \int_0^{-\ln \frac{1}{2}} \left[ \left( 1 - e^{-x} \right)^2 - \left( e^{-x} - 4.5 \right)^2 \right] \, dx }

We first expanded and simplified the expression inside the integral, and then evaluated the integral using the properties of definite integrals. Finally, we substituted the results into the original equation and simplified the result.

Introduction

In our previous article, we evaluated the volume of a solid using the given integral:

V=π∫0−ln⁡12[(1−e−x)2−(e−x−4.5)2] dx{ V = \pi \int_0^{-\ln \frac{1}{2}} \left[ \left( 1 - e^{-x} \right)^2 - \left( e^{-x} - 4.5 \right)^2 \right] \, dx }

We expanded and simplified the expression inside the integral, and then evaluated the integral using the properties of definite integrals. Finally, we substituted the results into the original equation and simplified the result.

In this article, we will answer some of the most frequently asked questions about evaluating the volume of a solid using a definite integral.

Q: What is the purpose of evaluating the volume of a solid using a definite integral?

A: The purpose of evaluating the volume of a solid using a definite integral is to calculate the volume of a solid that is defined by a given function. This is useful in a variety of fields, including physics, engineering, and mathematics.

Q: What are the steps involved in evaluating the volume of a solid using a definite integral?

A: The steps involved in evaluating the volume of a solid using a definite integral are:

  1. Expand and simplify the expression inside the integral.
  2. Evaluate the integral using the properties of definite integrals.
  3. Substitute the results into the original equation.
  4. Simplify the result.

Q: What are some common mistakes to avoid when evaluating the volume of a solid using a definite integral?

A: Some common mistakes to avoid when evaluating the volume of a solid using a definite integral include:

  • Failing to expand and simplify the expression inside the integral.
  • Failing to evaluate the integral using the properties of definite integrals.
  • Failing to substitute the results into the original equation.
  • Failing to simplify the result.

Q: What are some real-world applications of evaluating the volume of a solid using a definite integral?

A: Some real-world applications of evaluating the volume of a solid using a definite integral include:

  • Calculating the volume of a tank or a container.
  • Calculating the volume of a solid object, such as a sphere or a cylinder.
  • Calculating the volume of a region in space, such as a sphere or a cylinder.

Q: What are some tips for evaluating the volume of a solid using a definite integral?

A: Some tips for evaluating the volume of a solid using a definite integral include:

  • Make sure to expand and simplify the expression inside the integral.
  • Make sure to evaluate the integral using the properties of definite integrals.
  • Make sure to substitute the results into the original equation.
  • Make sure to simplify the result.

Q: What are some common challenges when evaluating the volume of a solid using a definite integral?

A: Some common challenges when evaluating the volume of a solid using a definite integral include:

  • Difficulty in expanding and simplifying the expression inside the integral.
  • Difficulty in evaluating the integral using the properties of definite integrals.
  • Difficulty in substituting the results into the original equation.
  • Difficulty in simplifying the result.

Conclusion

In this article, we answered some of the most frequently asked questions about evaluating the volume of a solid using a definite integral. We discussed the purpose of evaluating the volume of a solid using a definite integral, the steps involved in evaluating the volume of a solid using a definite integral, common mistakes to avoid, real-world applications, and tips for evaluating the volume of a solid using a definite integral. We also discussed some common challenges when evaluating the volume of a solid using a definite integral.

We hope that this article has been helpful in answering your questions about evaluating the volume of a solid using a definite integral. If you have any further questions, please don't hesitate to ask.