Evaluate The Integral:${ = \pi \int_0^{-\ln \left(\frac{12}{2}\right)}\left(-8+2 E^{-x}\right) Dx }$

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Introduction

In this article, we will delve into the world of calculus and evaluate the given integral. The integral in question is a definite integral, which means it has a specific upper and lower bound. Our goal is to find the value of this integral, and we will do so by applying various techniques from calculus.

The Integral

The given integral is:

∫0βˆ’ln⁑(122)(βˆ’8+2eβˆ’x)dx{ \int_0^{-\ln \left(\frac{12}{2}\right)}\left(-8+2 e^{-x}\right) dx }

This integral has a lower bound of 0 and an upper bound of βˆ’ln⁑(122)-\ln \left(\frac{12}{2}\right). We will need to evaluate this integral to find its value.

Evaluating the Integral

To evaluate this integral, we will first need to find the antiderivative of the function βˆ’8+2eβˆ’x-8+2 e^{-x}. The antiderivative of a function is a function whose derivative is the original function.

Finding the Antiderivative

The antiderivative of eβˆ’xe^{-x} is βˆ’eβˆ’x-e^{-x}. We can use this fact to find the antiderivative of 2eβˆ’x2 e^{-x}, which is βˆ’2eβˆ’x-2 e^{-x}.

However, we also have a constant term of βˆ’8-8 in the function. The antiderivative of a constant is the constant times the variable. In this case, the antiderivative of βˆ’8-8 is βˆ’8x-8x.

Combining the Antiderivatives

Now that we have found the antiderivatives of each term, we can combine them to find the antiderivative of the entire function.

∫(βˆ’8+2eβˆ’x)dx=βˆ’8xβˆ’2eβˆ’x+C{ \int\left(-8+2 e^{-x}\right) dx = -8x - 2 e^{-x} + C }

where CC is the constant of integration.

Applying the Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus states that the definite integral of a function can be evaluated by applying the antiderivative of the function to the upper and lower bounds of the integral.

In this case, we have:

∫0βˆ’ln⁑(122)(βˆ’8+2eβˆ’x)dx=[βˆ’8xβˆ’2eβˆ’x]0βˆ’ln⁑(122){ \int_0^{-\ln \left(\frac{12}{2}\right)}\left(-8+2 e^{-x}\right) dx = \left[-8x - 2 e^{-x}\right]_0^{-\ln \left(\frac{12}{2}\right)} }

Evaluating the Expression

Now that we have applied the Fundamental Theorem of Calculus, we can evaluate the expression.

[βˆ’8xβˆ’2eβˆ’x]0βˆ’ln⁑(122)=[βˆ’8(βˆ’ln⁑(122))βˆ’2eln⁑(122)]βˆ’[βˆ’8(0)βˆ’2e0]{ \left[-8x - 2 e^{-x}\right]_0^{-\ln \left(\frac{12}{2}\right)} = \left[-8\left(-\ln \left(\frac{12}{2}\right)\right) - 2 e^{\ln \left(\frac{12}{2}\right)}\right] - \left[-8(0) - 2 e^{0}\right] }

Simplifying the Expression

We can simplify the expression by evaluating the logarithm and the exponential functions.

[βˆ’8(βˆ’ln⁑(122))βˆ’2eln⁑(122)]βˆ’[βˆ’8(0)βˆ’2e0]=[βˆ’8(βˆ’ln⁑(6))βˆ’2(122)]βˆ’[βˆ’8(0)βˆ’2]{ \left[-8\left(-\ln \left(\frac{12}{2}\right)\right) - 2 e^{\ln \left(\frac{12}{2}\right)}\right] - \left[-8(0) - 2 e^{0}\right] = \left[-8\left(-\ln \left(6\right)\right) - 2 \left(\frac{12}{2}\right)\right] - \left[-8(0) - 2\right] }

Further Simplification

We can further simplify the expression by evaluating the logarithm and the exponential functions.

[βˆ’8(βˆ’ln⁑(6))βˆ’2(122)]βˆ’[βˆ’8(0)βˆ’2]=[8ln⁑(6)βˆ’12]βˆ’[βˆ’2]{ \left[-8\left(-\ln \left(6\right)\right) - 2 \left(\frac{12}{2}\right)\right] - \left[-8(0) - 2\right] = \left[8\ln \left(6\right) - 12\right] - \left[-2\right] }

Final Simplification

We can finally simplify the expression by combining the terms.

[8ln⁑(6)βˆ’12]βˆ’[βˆ’2]=8ln⁑(6)βˆ’10{ \left[8\ln \left(6\right) - 12\right] - \left[-2\right] = 8\ln \left(6\right) - 10 }

Conclusion

In this article, we evaluated the given integral using various techniques from calculus. We found the antiderivative of the function, applied the Fundamental Theorem of Calculus, and simplified the expression to find the final value of the integral.

The final value of the integral is 8ln⁑(6)βˆ’108\ln \left(6\right) - 10. This is the solution to the given problem.

Final Answer

The final answer is: 8ln⁑(6)βˆ’10\boxed{8\ln \left(6\right) - 10}

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Introduction

In our previous article, we evaluated the given integral using various techniques from calculus. In this article, we will answer some frequently asked questions related to the integral and provide additional insights.

Q&A

Q: What is the significance of the lower and upper bounds of the integral?

A: The lower and upper bounds of the integral are crucial in determining the value of the integral. In this case, the lower bound is 0 and the upper bound is βˆ’ln⁑(122)-\ln \left(\frac{12}{2}\right). These bounds help us to evaluate the integral and find its value.

Q: How do we find the antiderivative of a function?

A: To find the antiderivative of a function, we need to find a function whose derivative is the original function. In this case, we found the antiderivative of eβˆ’xe^{-x} to be βˆ’eβˆ’x-e^{-x} and the antiderivative of βˆ’8-8 to be βˆ’8x-8x.

Q: What is the Fundamental Theorem of Calculus?

A: The Fundamental Theorem of Calculus states that the definite integral of a function can be evaluated by applying the antiderivative of the function to the upper and lower bounds of the integral.

Q: How do we apply the Fundamental Theorem of Calculus?

A: To apply the Fundamental Theorem of Calculus, we need to evaluate the antiderivative of the function at the upper and lower bounds of the integral and then subtract the values.

Q: What is the final value of the integral?

A: The final value of the integral is 8ln⁑(6)βˆ’108\ln \left(6\right) - 10.

Q: Can we use other techniques to evaluate the integral?

A: Yes, we can use other techniques such as substitution or integration by parts to evaluate the integral. However, in this case, we used the Fundamental Theorem of Calculus to find the value of the integral.

Q: What is the significance of the constant of integration?

A: The constant of integration is a constant that is added to the antiderivative of the function. It is used to ensure that the antiderivative is correct.

Q: Can we use the integral to solve real-world problems?

A: Yes, we can use the integral to solve real-world problems. For example, we can use the integral to find the area under a curve or the volume of a solid.

Additional Insights

Using the Integral to Find the Area Under a Curve

We can use the integral to find the area under a curve. For example, if we have a function f(x)f(x) and we want to find the area under the curve from x=ax=a to x=bx=b, we can use the integral:

∫abf(x)dx{ \int_a^b f(x) dx }

This will give us the area under the curve.

Using the Integral to Find the Volume of a Solid

We can use the integral to find the volume of a solid. For example, if we have a function f(x)f(x) and we want to find the volume of the solid formed by rotating the curve about the x-axis from x=ax=a to x=bx=b, we can use the integral:

∫abΟ€(f(x))2dx{ \int_a^b \pi \left(f(x)\right)^2 dx }

This will give us the volume of the solid.

Conclusion

In this article, we answered some frequently asked questions related to the integral and provided additional insights. We discussed the significance of the lower and upper bounds of the integral, how to find the antiderivative of a function, and how to apply the Fundamental Theorem of Calculus. We also discussed the final value of the integral and how to use the integral to solve real-world problems.

Final Answer

The final answer is: 8ln⁑(6)βˆ’10\boxed{8\ln \left(6\right) - 10}