Use The Formula $\int U E^{a U} \, D U = \frac{1}{a^2}(a U - 1) E^{a U} + C$ To Find $\int X E^{5 X} \, D X$. Fill In The Box $\square$ With Your Answer And Add $+ C$.

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Introduction

In calculus, integrals of exponential functions are a common and essential topic. The formula ∫ueau du=1a2(auβˆ’1)eau+C\int u e^{a u} \, d u = \frac{1}{a^2}(a u - 1) e^{a u} + C is a powerful tool for solving these types of integrals. In this article, we will use this formula to find the integral of xe5xx e^{5 x}.

The Formula

The given formula is ∫ueau du=1a2(auβˆ’1)eau+C\int u e^{a u} \, d u = \frac{1}{a^2}(a u - 1) e^{a u} + C. This formula is a general solution for integrals of the form ∫ueau du\int u e^{a u} \, d u. To use this formula, we need to identify the values of uu and aa in the given integral.

Identifying the Values of uu and aa

In the given integral ∫xe5x dx\int x e^{5 x} \, d x, we can identify u=xu = x and a=5a = 5. Now that we have identified the values of uu and aa, we can substitute them into the formula.

Substituting the Values into the Formula

Substituting u=xu = x and a=5a = 5 into the formula, we get:

∫xe5x dx=152(5xβˆ’1)e5x+C\int x e^{5 x} \, d x = \frac{1}{5^2}(5 x - 1) e^{5 x} + C

Simplifying the Expression

Simplifying the expression, we get:

∫xe5x dx=125(5xβˆ’1)e5x+C\int x e^{5 x} \, d x = \frac{1}{25}(5 x - 1) e^{5 x} + C

The Final Answer

Therefore, the final answer is:

125(5xβˆ’1)e5x+C\boxed{\frac{1}{25}(5 x - 1) e^{5 x} + C}

Conclusion

In this article, we used the formula ∫ueau du=1a2(auβˆ’1)eau+C\int u e^{a u} \, d u = \frac{1}{a^2}(a u - 1) e^{a u} + C to find the integral of xe5xx e^{5 x}. We identified the values of uu and aa, substituted them into the formula, and simplified the expression to get the final answer.

Additional Examples

Here are a few additional examples of using the formula to solve integrals of exponential functions:

  • ∫xe2x dx=122(2xβˆ’1)e2x+C\int x e^{2 x} \, d x = \frac{1}{2^2}(2 x - 1) e^{2 x} + C
  • ∫xe3x dx=132(3xβˆ’1)e3x+C\int x e^{3 x} \, d x = \frac{1}{3^2}(3 x - 1) e^{3 x} + C
  • ∫xe4x dx=142(4xβˆ’1)e4x+C\int x e^{4 x} \, d x = \frac{1}{4^2}(4 x - 1) e^{4 x} + C

Tips and Tricks

Here are a few tips and tricks for using the formula to solve integrals of exponential functions:

  • Make sure to identify the values of uu and aa correctly.
  • Substitute the values into the formula carefully.
  • Simplify the expression to get the final answer.

Common Mistakes

Here are a few common mistakes to avoid when using the formula to solve integrals of exponential functions:

  • Not identifying the values of uu and aa correctly.
  • Not substituting the values into the formula carefully.
  • Not simplifying the expression to get the final answer.

Conclusion

Introduction

In our previous article, we used the formula ∫ueau du=1a2(auβˆ’1)eau+C\int u e^{a u} \, d u = \frac{1}{a^2}(a u - 1) e^{a u} + C to find the integral of xe5xx e^{5 x}. In this article, we will answer some frequently asked questions about using the formula to solve integrals of exponential functions.

Q: What is the formula for solving integrals of exponential functions?

A: The formula for solving integrals of exponential functions is ∫ueau du=1a2(auβˆ’1)eau+C\int u e^{a u} \, d u = \frac{1}{a^2}(a u - 1) e^{a u} + C.

Q: How do I identify the values of uu and aa in the given integral?

A: To identify the values of uu and aa, you need to look at the given integral and identify the function that is being integrated. In the case of ∫xe5x dx\int x e^{5 x} \, d x, u=xu = x and a=5a = 5.

Q: What if I have a different integral, such as ∫xe2x dx\int x e^{2 x} \, d x? How do I use the formula?

A: To use the formula for ∫xe2x dx\int x e^{2 x} \, d x, you need to identify the values of uu and aa. In this case, u=xu = x and a=2a = 2. Then, you can substitute these values into the formula and simplify the expression to get the final answer.

Q: What if I make a mistake when identifying the values of uu and aa? How do I correct it?

A: If you make a mistake when identifying the values of uu and aa, you need to go back and re-examine the given integral. Make sure to identify the function that is being integrated and the constant that is being multiplied by the function. Once you have correctly identified the values of uu and aa, you can substitute them into the formula and simplify the expression to get the final answer.

Q: Can I use the formula to solve integrals of other types of functions, such as trigonometric functions?

A: No, the formula ∫ueau du=1a2(auβˆ’1)eau+C\int u e^{a u} \, d u = \frac{1}{a^2}(a u - 1) e^{a u} + C is only for solving integrals of exponential functions. If you have an integral of a trigonometric function, you will need to use a different formula or technique to solve it.

Q: How do I know if I have used the formula correctly?

A: To know if you have used the formula correctly, you need to check your work carefully. Make sure to identify the values of uu and aa correctly, substitute them into the formula, and simplify the expression to get the final answer. If you have made a mistake, you will need to go back and re-examine your work.

Q: What if I get a different answer than the one in the textbook or online resource? How do I know which answer is correct?

A: If you get a different answer than the one in the textbook or online resource, you need to go back and re-examine your work. Make sure to identify the values of uu and aa correctly, substitute them into the formula, and simplify the expression to get the final answer. If you have made a mistake, you will need to correct it and try again.

Conclusion

In conclusion, the formula ∫ueau du=1a2(auβˆ’1)eau+C\int u e^{a u} \, d u = \frac{1}{a^2}(a u - 1) e^{a u} + C is a powerful tool for solving integrals of exponential functions. By identifying the values of uu and aa, substituting them into the formula, and simplifying the expression, we can find the integral of xe5xx e^{5 x}. We hope that this Q&A article has been helpful in answering some of the frequently asked questions about using the formula to solve integrals of exponential functions.