Problem 15 - Calculus

Question

Find the indefinite integrels in \(1-20\). \(\int(1+\sqrt{x}) d x / \sqrt{x}\)

Step-by-Step Solution

Answer

The indefinite integral is \(2\sqrt{x} + x + C\).

Step 1: Simplify the Integrand

First, simplify the expression \(\frac{1 + \sqrt{x}}{\sqrt{x}}\). We can separate this fraction into two terms: \(\frac{1}{\sqrt{x}} + \frac{\sqrt{x}}{\sqrt{x}}\). The second term simplifies to 1, so the integrand becomes \(x^{-1/2} + 1\).

Step 2: Integrate Each Term Separately

Now integrate each term: \(\int x^{-1/2} \, dx + \int 1 \, dx\).For the first term, use the power rule \(\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\), which gives: \(\int x^{-1/2} \, dx = \frac{x^{(-1/2) + 1}}{-1/2 + 1} + C = 2x^{1/2} + C_1\).For the second term, integrate directly: \(\int 1 \, dx = x + C_2\).

Step 3: Combine the Results

Add the results of the two integrals together to get the final answer.Combine the expressions:\(2x^{1/2} + x + C\), where \(C = C_1 + C_2\) is the constant of integration.

Key Concepts

Power Rule

The Power Rule is a fundamental technique in calculus for finding the indefinite integral of powers of x. It is especially useful when the function to be integrated is a monomial. The rule states that for any real number \(n eq -1\), the integral of \(x^n\) with respect to \(x\) is given by:\[ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \]where \(C\) is the constant of integration. This rule simplifies the process of integration by providing a straightforward formula to apply. In our exercise, we used the Power Rule to integrate the term \(x^{-1/2}\). We added one to the exponent \(-1/2\), and then divided by the new exponent \(1/2\), resulting in \(2x^{1/2}\). Note that the Power Rule only works for exponents other than \(-1\) because special techniques are needed for \(x^{-1}\), which results in the natural logarithm function.

Integration Techniques

Integration Techniques involve various methods to simplify and calculate integrals, especially when they can't be solved immediately. In this exercise, we employed the basic technique of separating terms in the integrand. Initially, the integrand \(\frac{1 + \sqrt{x}}{\sqrt{x}}\) was not easily integrable. By rewriting it as \(x^{-1/2} + 1\), we made it suitable for direct integration with the Power Rule.

• Separating the terms simplified the integrand, allowing us to apply integration techniques directly to each term.
• Each term can then be integrated separately: \(\int x^{-1/2} \, dx\) and \(\int 1 \, dx\).

This technique is often used when dealing with rational expressions or expressions that can be decomposed into simpler parts. Understanding how to manipulate and simplify expressions is crucial in making integration feasible.

Simplifying Expressions

Simplifying expressions is an essential skill in calculus that can make the process of integration far easier. In this problem, we initially had the complex fraction \(\frac{1 + \sqrt{x}}{\sqrt{x}}\). By separating this fraction into smaller, more manageable parts: \(\frac{1}{\sqrt{x}}\) and \(\frac{\sqrt{x}}{\sqrt{x}}\), we reduced it to \(x^{-1/2} + 1\). This simplification was crucial as it allowed us to use basic integration techniques.

• The second term simplifies directly to 1 because any number divided by itself equals 1.
• Simplification often involves converting roots to exponents, as seen with \(\sqrt{x} = x^{1/2}\), which aids in applying the Power Rule.

Remember, simplification can transform an intimidating integral into something much more straightforward to handle, especially when it involves algebraic fractions and roots.

Constant of Integration

The Constant of Integration is a crucial element in calculus to consider when finding indefinite integrals. When we integrate a function, we are essentially reversing the process of differentiation, which is why we must account for a possible constant term. This constant, denoted as \(C\), represents an infinite number of possible shifts up or down on a graph. In the solution process, each part of the integral gets its own constant (\(C_1, C_2\)), and these are combined at the end into a single constant of integration \(C\).

• In our exercise, the constants \(C_1\) and \(C_2\) from each integrated term were combined into a single constant \(C\) in the final expression.
• Even when omitted in intermediate steps, always include \(C\) in the final indefinite integral result to reflect all possible functions that could have been differentiated to get the integrand.

Ignoring the constant of integration in indefinite integrals would lead to incomplete expressions, losing the generality needed in calculus solutions.