Evaluate ∫ 0 4 ( 4 E 0.25 X + 4 X ) D X \int_0^4\left(4 E^{0.25 X}+4 X\right) D X ∫ 0 4 ​ ( 4 E 0.25 X + 4 X ) D X And Express The Answer In Simplest Form.

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Introduction

The given problem involves evaluating a definite integral, which is a fundamental concept in calculus. The integral in question is $\int_0^4\left(4 e^{0.25 x}+4 x\right) d x$. To evaluate this integral, we will use the properties of definite integrals and the fundamental theorem of calculus.

Breaking Down the Integral

The given integral can be broken down into two separate integrals:

$\int_0^4\left(4 e^{0.25 x}+4 x\right) d x = \int_0^4 4 e^{0.25 x} d x + \int_0^4 4 x d x$

Evaluating the First Integral

To evaluate the first integral, we will use the property of definite integrals that states:

$\int_0^a f(x) d x = F(a) - F(0)$

where $F(x)$ is the antiderivative of $f(x)$.

The antiderivative of $e^{0.25 x}$ is $\frac{4}{0.25} e^{0.25 x} = 16 e^{0.25 x}$.

Therefore, the first integral can be evaluated as:

$\int_0^4 4 e^{0.25 x} d x = 4 \left[16 e^{0.25 x}\right]_0^4 = 4 \left(16 e^{0.25 \cdot 4} - 16 e^{0.25 \cdot 0}\right)$

$= 4 \left(16 e^1 - 16 e^0\right) = 4 \left(16 e - 16\right) = 64 e - 64$

Evaluating the Second Integral

To evaluate the second integral, we will use the property of definite integrals that states:

$\int_0^a x d x = \frac{a^2}{2}$

Therefore, the second integral can be evaluated as:

$\int_0^4 4 x d x = 4 \left[\frac{x^2}{2}\right]_0^4 = 4 \left(\frac{4^2}{2} - \frac{0^2}{2}\right)$

$= 4 \left(\frac{16}{2} - 0\right) = 4 \cdot 8 = 32$

Combining the Results

Now that we have evaluated both integrals, we can combine the results to get the final answer:

$\int_0^4\left(4 e^{0.25 x}+4 x\right) d x = 64 e - 64 + 32$

$= 64 e - 32$

Conclusion

In this article, we evaluated the definite integral $\int_0^4\left(4 e^{0.25 x}+4 x\right) d x$ and expressed the answer in simplest form. We broke down the integral into two separate integrals, evaluated each integral using the properties of definite integrals, and combined the results to get the final answer.

Final Answer

The final answer is $\boxed{64 e - 32}$.

Step-by-Step Solution

Here is the step-by-step solution to the problem:

1. Break down the integral into two separate integrals: $\int_0^4 4 e^{0.25 x} d x + \int_0^4 4 x d x$
2. Evaluate the first integral using the property of definite integrals: $\int_0^a f(x) d x = F(a) - F(0)$
3. Find the antiderivative of $e^{0.25 x}$: $\frac{4}{0.25} e^{0.25 x} = 16 e^{0.25 x}$
4. Evaluate the first integral: $\int_0^4 4 e^{0.25 x} d x = 4 \left(16 e^{0.25 \cdot 4} - 16 e^{0.25 \cdot 0}\right) = 64 e - 64$
5. Evaluate the second integral using the property of definite integrals: $\int_0^a x d x = \frac{a^2}{2}$
6. Evaluate the second integral: $\int_0^4 4 x d x = 4 \left(\frac{4^2}{2} - \frac{0^2}{2}\right) = 32$
7. Combine the results to get the final answer: $\int_0^4\left(4 e^{0.25 x}+4 x\right) d x = 64 e - 64 + 32 = 64 e - 32$

Frequently Asked Questions

Here are some frequently asked questions related to the problem:

• Q: What is the final answer to the problem? A: The final answer is $\boxed{64 e - 32}$.
• Q: How do I evaluate the first integral? A: To evaluate the first integral, use the property of definite integrals: $\int_0^a f(x) d x = F(a) - F(0)$.
• Q: How do I find the antiderivative of $e^{0.25 x}$? A: The antiderivative of $e^{0.25 x}$ is $\frac{4}{0.25} e^{0.25 x} = 16 e^{0.25 x}$.
• Q: How do I evaluate the second integral? A: To evaluate the second integral, use the property of definite integrals: $\int_0^a x d x = \frac{a^2}{2}$.

Related Topics

Here are some related topics to the problem:

• Definite integrals
• Antiderivatives
• Properties of definite integrals
• Calculus

References

Here are some references related to the problem:

• [1] Calculus by Michael Spivak
• [2] Differential Equations by Lawrence Perko
• [3] Calculus: Early Transcendentals by James Stewart

Note: The references provided are for educational purposes only and are not intended to be a comprehensive list of resources.

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Introduction

In our previous article, we evaluated the definite integral $\int_0^4\left(4 e^{0.25 x}+4 x\right) d x$ and expressed the answer in simplest form. In this article, we will answer some frequently asked questions related to the problem.

Q&A

Q: What is the final answer to the problem?

A: The final answer is $\boxed{64 e - 32}$.

Q: How do I evaluate the first integral?

A: To evaluate the first integral, use the property of definite integrals: $\int_0^a f(x) d x = F(a) - F(0)$. Find the antiderivative of $e^{0.25 x}$, which is $\frac{4}{0.25} e^{0.25 x} = 16 e^{0.25 x}$. Then, evaluate the antiderivative at the limits of integration: $16 e^{0.25 \cdot 4} - 16 e^{0.25 \cdot 0}$.

Q: How do I find the antiderivative of $e^{0.25 x}$?

A: The antiderivative of $e^{0.25 x}$ is $\frac{4}{0.25} e^{0.25 x} = 16 e^{0.25 x}$.

Q: How do I evaluate the second integral?

A: To evaluate the second integral, use the property of definite integrals: $\int_0^a x d x = \frac{a^2}{2}$. Evaluate the expression at the limits of integration: $\frac{4^2}{2} - \frac{0^2}{2}$.

Q: What is the difference between a definite integral and an indefinite integral?

A: A definite integral is an integral with a specific upper and lower bound, while an indefinite integral is an integral without a specific upper and lower bound.

Q: How do I use the fundamental theorem of calculus to evaluate a definite integral?

A: The fundamental theorem of calculus states that the definite integral of a function $f(x)$ from $a$ to $b$ is equal to $F(b) - F(a)$, where $F(x)$ is the antiderivative of $f(x)$.

Q: What is the antiderivative of $x^2$?

A: The antiderivative of $x^2$ is $\frac{x^3}{3}$.

Q: How do I evaluate the definite integral $\int_0^2 x^2 d x$?

A: To evaluate the definite integral, use the property of definite integrals: $\int_0^a x^2 d x = \frac{a^3}{3}$. Evaluate the expression at the limits of integration: $\frac{2^3}{3} - \frac{0^3}{3}$.

Conclusion

In this article, we answered some frequently asked questions related to evaluating definite integrals. We covered topics such as the final answer to the problem, how to evaluate the first integral, how to find the antiderivative of $e^{0.25 x}$, and how to use the fundamental theorem of calculus to evaluate a definite integral.

Final Answer

The final answer is $\boxed{64 e - 32}$.

Step-by-Step Solution

Here is the step-by-step solution to the problem:

1. Break down the integral into two separate integrals: $\int_0^4 4 e^{0.25 x} d x + \int_0^4 4 x d x$
2. Evaluate the first integral using the property of definite integrals: $\int_0^a f(x) d x = F(a) - F(0)$
3. Find the antiderivative of $e^{0.25 x}$: $\frac{4}{0.25} e^{0.25 x} = 16 e^{0.25 x}$
4. Evaluate the first integral: $\int_0^4 4 e^{0.25 x} d x = 4 \left(16 e^{0.25 \cdot 4} - 16 e^{0.25 \cdot 0}\right) = 64 e - 64$
5. Evaluate the second integral using the property of definite integrals: $\int_0^a x d x = \frac{a^2}{2}$
6. Evaluate the second integral: $\int_0^4 4 x d x = 4 \left(\frac{4^2}{2} - \frac{0^2}{2}\right) = 32$
7. Combine the results to get the final answer: $\int_0^4\left(4 e^{0.25 x}+4 x\right) d x = 64 e - 64 + 32 = 64 e - 32$

Frequently Asked Questions

Here are some frequently asked questions related to the problem:

• Q: What is the final answer to the problem? A: The final answer is $\boxed{64 e - 32}$.
• Q: How do I evaluate the first integral? A: To evaluate the first integral, use the property of definite integrals: $\int_0^a f(x) d x = F(a) - F(0)$.
• Q: How do I find the antiderivative of $e^{0.25 x}$? A: The antiderivative of $e^{0.25 x}$ is $\frac{4}{0.25} e^{0.25 x} = 16 e^{0.25 x}$.
• Q: How do I evaluate the second integral? A: To evaluate the second integral, use the property of definite integrals: $\int_0^a x d x = \frac{a^2}{2}$.

Related Topics

Here are some related topics to the problem:

• Definite integrals
• Antiderivatives
• Properties of definite integrals
• Calculus

References

Here are some references related to the problem:

• [1] Calculus by Michael Spivak
• [2] Differential Equations by Lawrence Perko
• [3] Calculus: Early Transcendentals by James Stewart

Note: The references provided are for educational purposes only and are not intended to be a comprehensive list of resources.