Evaluate The Definite Integral Of 1/(√(x+9) - √x) From 0 To 16

This article explores the process of evaluating the definite integral of the function ${ f(x) = \frac{1}{\sqrt{x+9} - \sqrt{x}} }$ from 0 to 16. This type of integral often appears in calculus and requires a clever approach to simplify the integrand before standard integration techniques can be applied. We'll delve into the steps required to rationalize the denominator, find the antiderivative, and finally, compute the definite integral.

Understanding the Integral

The integral we aim to solve is:

${ \int_0^{16} \frac{dx}{\sqrt{x+9} - \sqrt{x}} }$

The integrand, ${ \frac{1}{\sqrt{x+9} - \sqrt{x}} }$, presents a challenge because of the square roots in the denominator. Directly integrating this expression is not straightforward. Therefore, our initial focus will be on simplifying this expression through a technique known as rationalizing the denominator. This involves multiplying both the numerator and the denominator by the conjugate of the denominator.

Rationalizing the Denominator: A Crucial First Step

To rationalize the denominator, we multiply both the numerator and the denominator by the conjugate of ${ \sqrt{x+9} - \sqrt{x} }$, which is ${ \sqrt{x+9} + \sqrt{x} }$. This process eliminates the square roots from the denominator, making the integrand much easier to handle. The key here is the difference of squares identity: ${ (a - b)(a + b) = a^2 - b^2 }$.

Applying this, we get:

${ \frac{1}{\sqrt{x+9} - \sqrt{x}} \cdot \frac{\sqrt{x+9} + \sqrt{x}}{\sqrt{x+9} + \sqrt{x}} = \frac{\sqrt{x+9} + \sqrt{x}}{(x+9) - x} }$

Simplifying the denominator, we have:

${ \frac{\sqrt{x+9} + \sqrt{x}}{9} }$

Thus, the integral now becomes:

${ \int_0^{16} \frac{\sqrt{x+9} + \sqrt{x}}{9} dx = \frac{1}{9} \int_0^{16} (\sqrt{x+9} + \sqrt{x}) dx }$

This simplified form is much more amenable to integration. We have successfully transformed a complex integrand into a sum of two simpler square root functions, each of which can be integrated using the power rule for integration.

Integrating the Simplified Expression

Now that we have a simplified integrand, we can proceed with integration. The integral is now expressed as:

${ \frac{1}{9} \int_0^{16} (\sqrt{x+9} + \sqrt{x}) dx }$

We can split this integral into two separate integrals:

${ \frac{1}{9} \left[ \int_0^{16} \sqrt{x+9} dx + \int_0^{16} \sqrt{x} dx \right] }$

Each of these integrals can be solved using the power rule for integration. Recall that the power rule states:

${ \int x^n dx = \frac{x^{n+1}}{n+1} + C }$

where ${ n \neq -1 }$ and ${ C }$ is the constant of integration. For the first integral, ${ \int_0^{16} \sqrt{x+9} dx }$, we can use a simple substitution. Let ${ u = x + 9 }$, so ${ du = dx }$. The limits of integration change accordingly: when ${ x = 0 }$, ${ u = 9 }$, and when ${ x = 16 }$, ${ u = 25 }$. Thus, the first integral becomes:

${ \int_9^{25} \sqrt{u} du = \int_9^{25} u^{\frac{1}{2}} du }$

Applying the power rule, we get:

${ \left[ \frac{2}{3} u^{\frac{3}{2}} \right]_9^{25} = \frac{2}{3} (25^{\frac{3}{2}} - 9^{\frac{3}{2}}) = \frac{2}{3} (125 - 27) = \frac{2}{3} (98) = \frac{196}{3} }$

For the second integral, ${ \int_0^{16} \sqrt{x} dx }$, we directly apply the power rule:

${ \int_0^{16} x^{\frac{1}{2}} dx = \left[ \frac{2}{3} x^{\frac{3}{2}} \right]_0^{16} = \frac{2}{3} (16^{\frac{3}{2}} - 0^{\frac{3}{2}}) = \frac{2}{3} (64) = \frac{128}{3} }$

Now, we combine these results and multiply by ${ \frac{1}{9} }$:

${ \frac{1}{9} \left[ \frac{196}{3} + \frac{128}{3} \right] = \frac{1}{9} \left[ \frac{324}{3} \right] = \frac{1}{9} (108) = 12 }$

Therefore, the definite integral evaluates to 12.

Step-by-Step Solution

1. Rationalize the denominator: Multiply the numerator and denominator by the conjugate of ${ \sqrt{x+9} - \sqrt{x} }$, which is ${ \sqrt{x+9} + \sqrt{x} }$.

   ${ \frac{1}{\sqrt{x+9} - \sqrt{x}} \cdot \frac{\sqrt{x+9} + \sqrt{x}}{\sqrt{x+9} + \sqrt{x}} = \frac{\sqrt{x+9} + \sqrt{x}}{9} }$

2. Rewrite the integral: Substitute the simplified integrand back into the integral.

   ${ \int_0^{16} \frac{dx}{\sqrt{x+9} - \sqrt{x}} = \frac{1}{9} \int_0^{16} (\sqrt{x+9} + \sqrt{x}) dx }$

3. Split the integral: Separate the integral into two parts.

   ${ \frac{1}{9} \left[ \int_0^{16} \sqrt{x+9} dx + \int_0^{16} \sqrt{x} dx \right] }$

4. Integrate ${ \sqrt{x+9} }$: Use the substitution ${ u = x + 9 }$.

   ${ \int_0^{16} \sqrt{x+9} dx = \int_9^{25} \sqrt{u} du = \frac{196}{3} }$

5. Integrate ${ \sqrt{x} }$:

   ${ \int_0^{16} \sqrt{x} dx = \frac{128}{3} }$

6. Combine the results: Add the results of the two integrals and multiply by ${ \frac{1}{9} }$.

   ${ \frac{1}{9} \left[ \frac{196}{3} + \frac{128}{3} \right] = 12 }$

Common Mistakes and How to Avoid Them

When evaluating definite integrals of this type, several common mistakes can occur. Being aware of these pitfalls can help ensure accuracy and efficiency in your calculations.

Forgetting to Rationalize the Denominator

One of the most frequent errors is attempting to integrate the function directly without first rationalizing the denominator. As we've seen, the integrand ${ \frac{1}{\sqrt{x+9} - \sqrt{x}} }$ is not easily integrable in its original form. By failing to rationalize, you'll likely find yourself stuck with a complex expression that resists standard integration techniques. Always remember to start by multiplying the numerator and denominator by the conjugate to simplify the expression.

Incorrectly Applying the Power Rule

The power rule is a fundamental tool for integration, but it's essential to apply it correctly. A common mistake is to forget to add 1 to the exponent and divide by the new exponent. For example, the integral of ${ \sqrt{x} }$ (which is ${ x^{\frac{1}{2}} }$) should be ${ \frac{2}{3}x^{\frac{3}{2}} }$, not ${ \frac{1}{2}x^{\frac{3}{2}} }$ or something else. Double-check your application of the power rule, especially when dealing with fractional exponents.

Neglecting the Chain Rule (or U-Substitution)

When integrating composite functions, such as ${ \sqrt{x+9} }$, it's crucial to consider the chain rule or use u-substitution. In this case, letting ${ u = x + 9 }$ simplifies the integral to ${ \int \sqrt{u} du }$, which is easily handled by the power rule. Forgetting this step will lead to an incorrect antiderivative.

Errors in Changing Limits of Integration

If you use u-substitution to evaluate a definite integral, remember to change the limits of integration to reflect the new variable. In our example, when substituting ${ u = x + 9 }$, the limits changed from ${ x = 0 }$ and ${ x = 16 }$ to ${ u = 9 }$ and ${ u = 25 }$, respectively. Failing to update the limits will result in evaluating the antiderivative at the wrong points, leading to an incorrect final answer.

Arithmetic Errors

Simple arithmetic mistakes can derail the entire process, especially when dealing with fractions and exponents. Take extra care when simplifying expressions, evaluating powers, and combining terms. It's often helpful to perform calculations in multiple steps and double-check each step to minimize the risk of errors.

Forgetting to Distribute and Simplify

After rationalizing the denominator, it’s crucial to distribute and simplify the resulting expression. In our case, we multiplied by the conjugate and simplified to get ${ \frac{\sqrt{x+9} + \sqrt{x}}{9} }$. Forgetting to divide by the constant or further simplify can complicate the integration process.

Not Double-Checking the Final Answer

Finally, it's always a good practice to double-check your final answer. If possible, use a calculator or computer algebra system to verify your result. This can catch any subtle errors that may have slipped through the manual calculations.

By being mindful of these common mistakes and taking the time to avoid them, you can improve your accuracy and confidence in evaluating definite integrals.

Conclusion

In summary, the definite integral ${ \int_0^{16} \frac{dx}{\sqrt{x+9} - \sqrt{x}} }$ is evaluated by first rationalizing the denominator, which simplifies the integrand to ${ \frac{\sqrt{x+9} + \sqrt{x}}{9} }$. We then split the integral into two parts, apply the power rule and u-substitution where necessary, and combine the results. The final value of the integral is 12. This exercise demonstrates the importance of algebraic manipulation and the proper application of integration techniques in solving calculus problems. Remember to always double-check your work and be mindful of common errors to ensure accuracy.