Evaluate The Integral:$\int_{-\frac{1}{3}}^0(3x+1)^{29} \, Dx$

Introduction

In this article, we will delve into the world of calculus and evaluate the integral of a high-power polynomial function. The given integral is $\int_{-\frac{1}{3}}^0(3x+1)^{29} \, dx$. This type of integral can be challenging to solve, but with the right techniques and strategies, we can break it down and find the solution.

Understanding the Integral

The given integral is a definite integral, which means it has a specific upper and lower bound. In this case, the lower bound is $-\frac{1}{3}$ and the upper bound is $0$. The integrand is a high-power polynomial function of the form $(3x+1)^{29}$.

Substitution Method

One of the most common techniques used to evaluate integrals is the substitution method. This method involves substituting a new variable into the integral to simplify it. In this case, we can substitute $u = 3x + 1$. This will allow us to rewrite the integral in terms of $u$.

Derivation of the Substitution

Let's derive the substitution by finding the derivative of $u$ with respect to $x$. We have:

$$\frac{du}{dx} = 3$$

This means that $du = 3dx$. We can now substitute this into the integral:

$$\int_{-\frac{1}{3}}^0(3x+1)^{29} \, dx = \int_{u(-\frac{1}{3})}^{u(0)} u^{29} \, \frac{du}{3}$$

Simplifying the Integral

Now that we have substituted $u$ into the integral, we can simplify it. We have:

$$\int_{u(-\frac{1}{3})}^{u(0)} u^{29} \, \frac{du}{3} = \frac{1}{3} \int_{u(-\frac{1}{3})}^{u(0)} u^{29} \, du$$

Evaluating the Integral

Now that we have simplified the integral, we can evaluate it. We have:

$$\frac{1}{3} \int_{u(-\frac{1}{3})}^{u(0)} u^{29} \, du = \frac{1}{3} \left[ \frac{u^{30}}{30} \right]_{u(-\frac{1}{3})}^{u(0)}$$

Finding the Limits of Integration

Now that we have evaluated the integral, we need to find the limits of integration. We have:

$$u(-\frac{1}{3}) = 3(-\frac{1}{3}) + 1 = 0$$

$$u(0) = 3(0) + 1 = 1$$

Substituting the Limits of Integration

Now that we have found the limits of integration, we can substitute them into the integral:

$$\frac{1}{3} \left[ \frac{u^{30}}{30} \right]_{u(0)}^{u(1)} = \frac{1}{3} \left[ \frac{1^{30}}{30} - \frac{0^{30}}{30} \right]$$

Simplifying the Expression

Now that we have substituted the limits of integration, we can simplify the expression:

$$\frac{1}{3} \left[ \frac{1^{30}}{30} - \frac{0^{30}}{30} \right] = \frac{1}{3} \left[ \frac{1}{30} - 0 \right]$$

Final Answer

The final answer is:

$$\frac{1}{3} \left[ \frac{1}{30} - 0 \right] = \frac{1}{90}$$

Conclusion

In this article, we evaluated the integral of a high-power polynomial function using the substitution method. We substituted $u = 3x + 1$ into the integral and simplified it. We then evaluated the integral and found the final answer to be $\frac{1}{90}$. This type of integral can be challenging to solve, but with the right techniques and strategies, we can break it down and find the solution.

References

• [1] Calculus, 3rd edition, Michael Spivak
• [2] Calculus, 2nd edition, James Stewart

Additional Resources

• [1] Khan Academy: Calculus
• [2] MIT OpenCourseWare: Calculus

Discussion

Introduction

In our previous article, we evaluated the integral of a high-power polynomial function using the substitution method. In this article, we will answer some of the most frequently asked questions about this topic.

Q: What is the substitution method?

A: The substitution method is a technique used to simplify integrals by substituting a new variable into the integral. This can help to make the integral easier to evaluate.

Q: Why is the substitution method useful?

A: The substitution method is useful because it can help to simplify complex integrals and make them easier to evaluate. It can also help to identify patterns and relationships between different functions.

Q: How do I choose the substitution?

A: Choosing the right substitution is an important part of the substitution method. You should choose a substitution that simplifies the integral and makes it easier to evaluate. In the case of the integral we evaluated earlier, we chose the substitution $u = 3x + 1$ because it simplified the integral and made it easier to evaluate.

Q: What are some common substitutions?

A: Some common substitutions include:

• $u = f(x)$, where $f(x)$ is a function that simplifies the integral
• $u = ax + b$, where $a$ and $b$ are constants that simplify the integral
• $u = \frac{1}{x}$, where $x$ is a variable that simplifies the integral

Q: How do I evaluate the integral after substitution?

A: After substituting the new variable into the integral, you should evaluate the integral using the new variable. In the case of the integral we evaluated earlier, we evaluated the integral using the variable $u$.

Q: What are some common mistakes to avoid when using the substitution method?

A: Some common mistakes to avoid when using the substitution method include:

• Not choosing the right substitution
• Not simplifying the integral enough
• Not evaluating the integral correctly

Q: Can the substitution method be used with other types of integrals?

A: Yes, the substitution method can be used with other types of integrals, including:

• Definite integrals
• Indefinite integrals
• Improper integrals

Q: Are there any other techniques for evaluating integrals?

A: Yes, there are other techniques for evaluating integrals, including:

• Integration by parts
• Integration by partial fractions
• Integration by trigonometric substitution

Conclusion

In this article, we answered some of the most frequently asked questions about the substitution method for evaluating integrals. We discussed the importance of choosing the right substitution, simplifying the integral, and evaluating the integral correctly. We also discussed some common mistakes to avoid and other techniques for evaluating integrals.

References

• [1] Calculus, 3rd edition, Michael Spivak
• [2] Calculus, 2nd edition, James Stewart

Additional Resources

• [1] Khan Academy: Calculus
• [2] MIT OpenCourseWare: Calculus

Discussion

This article discussed the substitution method for evaluating integrals. If you have any questions or would like to discuss this topic further, please leave a comment below.