Use A Substitution To Express The Integrand As A Rational Function, And Then Evaluate The Integral:$\int_{36}^{64} \frac{\sqrt{x}}{x-4} \, Dx$

Introduction

In this article, we will explore the process of using a substitution to express the integrand as a rational function and then evaluate the integral. This technique is a powerful tool in calculus, allowing us to simplify complex integrals and make them more manageable.

The Problem

The given integral is $\int_{36}^{64} \frac{\sqrt{x}}{x-4} \, dx$. Our goal is to express the integrand as a rational function and then evaluate the integral.

Step 1: Identify a Suitable Substitution

To express the integrand as a rational function, we need to identify a suitable substitution. In this case, we can let $u = \sqrt{x}$. This substitution will simplify the integrand and make it easier to evaluate.

Step 2: Express the Integrand as a Rational Function

Using the substitution $u = \sqrt{x}$, we can express the integrand as a rational function. We have:

$$\frac{\sqrt{x}}{x-4} = \frac{u}{u^2-16}$$

This is a rational function in terms of $u$, and we can now evaluate the integral.

Step 3: Evaluate the Integral

To evaluate the integral, we need to find the antiderivative of the rational function. We can do this by using the method of partial fractions.

Let's assume that the rational function can be written as:

$$\frac{u}{u^2-16} = \frac{A}{u-4} + \frac{B}{u+4}$$

where $A$ and $B$ are constants. We can then find the values of $A$ and $B$ by equating the numerator of the rational function to the sum of the two partial fractions.

After finding the values of $A$ and $B$, we can integrate the rational function to get:

$$\int \frac{u}{u^2-16} \, du = \frac{1}{8} \ln \left| \frac{u-4}{u+4} \right| + C$$

where $C$ is the constant of integration.

Step 4: Apply the Fundamental Theorem of Calculus

Now that we have found the antiderivative of the rational function, we can apply the Fundamental Theorem of Calculus to evaluate the integral.

We have:

$$\int_{36}^{64} \frac{\sqrt{x}}{x-4} \, dx = \left[ \frac{1}{8} \ln \left| \frac{u-4}{u+4} \right| \right]_{36}^{64}$$

Evaluating the antiderivative at the limits of integration, we get:

$$\int_{36}^{64} \frac{\sqrt{x}}{x-4} \, dx = \frac{1}{8} \ln \left| \frac{\sqrt{64}-4}{\sqrt{64}+4} \right| - \frac{1}{8} \ln \left| \frac{\sqrt{36}-4}{\sqrt{36}+4} \right|$$

Simplifying the expression, we get:

$$\int_{36}^{64} \frac{\sqrt{x}}{x-4} \, dx = \frac{1}{8} \ln \left| \frac{4}{12} \right| - \frac{1}{8} \ln \left| \frac{2}{8} \right|$$

$$\int_{36}^{64} \frac{\sqrt{x}}{x-4} \, dx = \frac{1}{8} \ln \left| \frac{1}{3} \right| - \frac{1}{8} \ln \left| \frac{1}{4} \right|$$

$$\int_{36}^{64} \frac{\sqrt{x}}{x-4} \, dx = \frac{1}{8} \ln \left| \frac{1}{3} \right| + \frac{1}{8} \ln \left| 4 \right|$$

$$\int_{36}^{64} \frac{\sqrt{x}}{x-4} \, dx = \frac{1}{8} \ln \left| \frac{4}{3} \right|$$

Therefore, the value of the integral is $\frac{1}{8} \ln \left| \frac{4}{3} \right|$.

Conclusion

In this article, we used a substitution to express the integrand as a rational function and then evaluated the integral. We identified a suitable substitution, expressed the integrand as a rational function, and then evaluated the integral using the method of partial fractions. Finally, we applied the Fundamental Theorem of Calculus to evaluate the integral. The value of the integral is $\frac{1}{8} \ln \left| \frac{4}{3} \right|$.

References

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

Glossary

• Integrand: The function being integrated.
• Rational function: A function that can be expressed as the ratio of two polynomials.
• Partial fractions: A method of expressing a rational function as the sum of simpler fractions.
• Fundamental Theorem of Calculus: A theorem that relates the derivative of a function to the definite integral of the function.
  Q&A: Using a Substitution to Express the Integrand as a Rational Function and Evaluate the Integral =============================================================================================

Introduction

In our previous article, we explored the process of using a substitution to express the integrand as a rational function and then evaluate the integral. In this article, we will answer some common questions related to this topic.

Q: What is a substitution in calculus?

A substitution in calculus is a technique used to simplify a complex integral by replacing the variable of integration with a new variable. This new variable is often a function of the original variable, and it is used to express the integrand in a simpler form.

Q: Why do we need to use a substitution to express the integrand as a rational function?

We need to use a substitution to express the integrand as a rational function because it allows us to simplify the integral and make it more manageable. By expressing the integrand as a rational function, we can use the method of partial fractions to evaluate the integral.

Q: What is the method of partial fractions?

The method of partial fractions is a technique used to express a rational function as the sum of simpler fractions. This is done by factoring the denominator of the rational function and then expressing it as the sum of simpler fractions.

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

To apply the Fundamental Theorem of Calculus, we need to find the antiderivative of the rational function and then evaluate it at the limits of integration. This will give us the value of the integral.

Q: What are some common substitutions used in calculus?

Some common substitutions used in calculus include:

• u-substitution: This is a substitution where we let u = f(x) and then express the integrand in terms of u.
• trigonometric substitution: This is a substitution where we let x = f(t) and then express the integrand in terms of t.
• hyperbolic substitution: This is a substitution where we let x = f(u) and then express the integrand in terms of u.

Q: How do we choose the right substitution for a given integral?

To choose the right substitution for a given integral, we need to consider the form of the integrand and the limits of integration. We should look for a substitution that will simplify the integrand and make it easier to evaluate.

Q: What are some common mistakes to avoid when using a substitution to express the integrand as a rational function?

Some common mistakes to avoid when using a substitution to express the integrand as a rational function include:

• Not checking the limits of integration: Make sure to check the limits of integration to ensure that the substitution is valid.
• Not simplifying the integrand: Make sure to simplify the integrand before evaluating the integral.
• Not using the correct method of partial fractions: Make sure to use the correct method of partial fractions to express the rational function as the sum of simpler fractions.

Conclusion

In this article, we answered some common questions related to using a substitution to express the integrand as a rational function and evaluate the integral. We discussed the importance of choosing the right substitution, the method of partial fractions, and the Fundamental Theorem of Calculus. We also highlighted some common mistakes to avoid when using a substitution to express the integrand as a rational function.

References

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

Glossary

• Integrand: The function being integrated.
• Rational function: A function that can be expressed as the ratio of two polynomials.
• Partial fractions: A method of expressing a rational function as the sum of simpler fractions.
• Fundamental Theorem of Calculus: A theorem that relates the derivative of a function to the definite integral of the function.
• Substitution: A technique used to simplify a complex integral by replacing the variable of integration with a new variable.