Integral Of X Sqrt X 1

The Definite and Indefinite Integral of x√(x+1): A Comprehensive Guide

The integral ∫x√(x+1) dx presents a seemingly straightforward challenge, yet it requires a careful application of integration techniques. This comprehensive guide will delve into both the indefinite and definite integrals of this function, providing a step-by-step solution, exploring alternative approaches, and offering insights into the broader context of integration within calculus.

Understanding the Problem: ∫x√(x+1) dx

Our goal is to find a function whose derivative is x√(x+1). This is not a simple power rule integration, and requires a strategic substitution to simplify the integrand. Direct integration is not feasible; we need to employ a substitution method to make it solvable. The key lies in strategically choosing a substitution that simplifies the expression within the square root.

Method 1: u-Substitution

The most effective approach is u-substitution. Let's define:

u = x + 1

This implies:

x = u - 1 and du = dx

Now, substitute these expressions into our original integral:

∫x√(x+1) dx = ∫(u - 1)√u du

This simplifies to:

∫(u^(3/2) - u^(1/2)) du

Now, we can apply the power rule of integration:

∫u^n du = (u^(n+1))/(n+1) + C (where C is the constant of integration)

Applying the power rule to each term:

(u^(5/2))/(5/2) - (u^(3/2))/(3/2) + C

Simplifying further:

(2/5)u^(5/2) - (2/3)u^(3/2) + C

Finally, substitute back x + 1 for u:

(2/5)(x + 1)^(5/2) - (2/3)(x + 1)^(3/2) + C

This is the indefinite integral of x√(x+1). The constant of integration, C, accounts for the family of functions that all have x√(x+1) as their derivative.

Method 2: Integration by Parts (A Less Efficient Approach)

While u-substitution is the most efficient method, integration by parts can also be applied, although it leads to a more complex process. Let's explore this alternative approach:

Integration by parts follows the formula: ∫u dv = uv - ∫v du

Let's choose:

• u = x => du = dx
• dv = √(x+1) dx => v = (2/3)(x+1)^(3/2) (This requires a simple u-substitution within the integration by parts itself)

Applying the integration by parts formula:

∫x√(x+1) dx = x[(2/3)(x+1)^(3/2)] - ∫[(2/3)(x+1)^(3/2)]dx

This integral is still not trivial, requiring further application of the power rule:

∫[(2/3)(x+1)^(3/2)]dx = (2/3) * [(2/5)(x+1)^(5/2)] + C' (where C' is another constant of integration)

Putting it all together:

(2/3)x(x+1)^(3/2) - (4/15)(x+1)^(5/2) + C'

While seemingly different, this result is equivalent to the solution obtained through u-substitution. The constants C and C' will absorb any differences that might appear to be present between the two forms. The u-substitution method is undeniably cleaner and more efficient.

Evaluating the Definite Integral

To evaluate the definite integral, we need limits of integration, say from 'a' to 'b'. The process involves substituting these limits into the indefinite integral and subtracting the result for the lower limit from the result for the upper limit.

Let's say we want to evaluate:

∫[a to b] x√(x+1) dx

Using the indefinite integral we derived earlier:

[(2/5)(x + 1)^(5/2) - (2/3)(x + 1)^(3/2)] [from a to b]

= [(2/5)(b + 1)^(5/2) - (2/3)(b + 1)^(3/2)] - [(2/5)(a + 1)^(5/2) - (2/3)(a + 1)^(3/2)]

This will yield a numerical result based on the values of 'a' and 'b'. Remember to always substitute the upper limit first and then subtract the result obtained by substituting the lower limit.

Applications and Further Exploration

The integral ∫x√(x+1) dx, while seemingly abstract, finds applications in various areas of mathematics and science where functions with similar structures arise:

• Calculus: Understanding this integral strengthens your proficiency in applying u-substitution and integration by parts, crucial techniques in advanced calculus.
• Physics: This type of integral might appear when calculating work done by a variable force, or when dealing with certain types of curves in kinematics.
• Engineering: Similar integrals are common in problems involving areas under curves, volumes of solids of revolution, or modeling various physical phenomena.

Exploring Variations:

Once you master this integral, you can tackle similar integrals with slight variations, for example:

• ∫x√(ax + b) dx (where a and b are constants) - a simple generalization requiring a slightly adjusted u-substitution.
• ∫x²√(x+1) dx - requiring a more complex u-substitution or perhaps integration by parts.
• ∫(x+k)√(x+1)dx - another simple variation that builds upon the core concepts.

Common Mistakes to Avoid

• Incorrect u-substitution: Choosing an inappropriate substitution can lead to a more complicated integral rather than a simplified one. Carefully choose your 'u' to simplify the expression.
• Forgetting the constant of integration (C): This is crucial for indefinite integrals, as it represents the entire family of antiderivatives.
• Errors in algebraic manipulation: Ensure accuracy when simplifying expressions and applying the power rule.
• Incorrect application of integration by parts: Make sure to follow the formula correctly and handle the remaining integral effectively.
• Order of operations when evaluating definite integrals: Remember to substitute the limits of integration correctly, subtracting the lower limit result from the upper limit result.

Conclusion

Mastering the integral ∫x√(x+1) dx not only enhances your calculus skills but also provides a foundation for tackling more complex integration problems. Remember the efficiency of u-substitution and the importance of meticulous algebraic manipulation. The ability to confidently solve this integral is a testament to your growing understanding of integration techniques and their broader applications in various fields. Regular practice and a systematic approach will solidify your mastery of these crucial concepts within calculus.