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Chapter 11: Problem 35

Find the indefinite integral. $$\int\left(\sqrt{x}+\frac{3}{\sqrt{x}}\right) d x$$

Short Answer

Expert verified
The short answer to the question is: \(\frac{2}{3}x^{\frac{3}{2}} + 6x^{\frac{1}{2}} + C\), where C is the constant of integration.

Step by step solution

01

Integrate Each Term Separately

First, we will rewrite the given function into two separate integrals: \( \int \sqrt{x}\,dx + \int \frac{3}{\sqrt{x}}\,dx \)
02

Rewrite the Terms

We will rewrite the square root as a fractional exponent. The square root of x can be written as \(x^{\frac{1}{2}}\) and \(x^{-\frac{1}{2}}\) for the 3 divided by the square root of x. The new expression looks like: \( \int x^{\frac{1}{2}}\,dx + \int 3x^{-\frac{1}{2}}\,dx \)
03

Apply Power Rule for Integration

Now, let's integrate each term using the power rule for integration, which states that \( \int x^a\,dx = \frac{x^{a+1}}{a+1} + C \), where a is the exponent. For the first term, the power rule gives us: \( \frac{x^{\frac{1}{2} + 1}}{\frac{1}{2} + 1} + C_1\) Which simplifies to: \( \frac{2}{3}x^{\frac{3}{2}} + C_1 \) For the second term, we get: \( 3\frac{x^{-\frac{1}{2} + 1}}{-\frac{1}{2} + 1} + C_2\) Which simplifies to: \( 6x^{\frac{1}{2}} + C_2 \)
04

Combine the Integrals

Now we combine the terms to get the indefinite integral of the given function: \( \frac{2}{3}x^{\frac{3}{2}} + 6x^{\frac{1}{2}} + C \) where C is the constant of integration, and C = C_1 + C_2.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Power Rule for Integration
The power rule for integration is a very handy technique. It allows us to find the integral of functions that have terms of the form \(x^a\).
In the context of indefinite integrals, this rule tells us:
  • If you want to integrate \(x^a\), you increase the exponent by one, and then divide by the new exponent.
  • It's written mathematically as \( \int x^a \, dx = \frac{x^{a+1}}{a+1} + C\).
  • Here, \(a\) is any real number, and \(C\) is the constant of integration.
  • Remember, this rule doesn't work for \(a = -1\), because then you're dividing by zero!
For example, in our exercise, we applied the power rule to \(x^{\frac{1}{2}}\) and \(x^{-\frac{1}{2}}\). The integration changes their powers, helping us solve the problem effortlessly.
Fractional Exponents
In mathematics, understanding fractional exponents is crucial. They allow us to express roots more generally.
For instance, the square root \(\sqrt{x}\) can be written as \(x^{\frac{1}{2}}\).
This becomes particularly useful for integration because it brings roots into a form where the power rule can be applied. Here's why fractional exponents matter:
  • They simplify operations, especially when integrating or differentiating functions involving roots.
  • They turn complicated root expressions into straightforward terms that follow exponent rules such as multiplication and division.
In our example, both \(\sqrt{x}\) and \(\frac{1}{\sqrt{x}}\) are rewritten using fractional exponents, which were \(x^{\frac{1}{2}}\) and \(x^{-\frac{1}{2}}\) respectively.
Constant of Integration
Whenever we calculate indefinite integrals, we must include a constant of integration. This is because an indefinite integral represents a family of functions.
Here鈥檚 why the constant is important:
  • The derivative of any constant is zero. When you integrate, you're essentially reversing differentiation, but you cannot tell if there was a constant initially.
  • This means each antiderivative differs by a constant, which we represent as \(C\).
  • Including \(C\) ensures the integration solution is correct, covering all possible original functions.
In the solution, we see \(C = C_1 + C_2\). Each integration step adds its own \(C\), and together they form the full constant for the entire integral.
Remember, this constant may seem minor but is essential for completing the solution.

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Most popular questions from this chapter

Sketch the graphs of the functions \(f\) and \(g\) and find the area of the region enclosed by these graphs and the vertical lines \(x=a\) and \(x=b\). $$f(x)=e^{x}, g(x)=\frac{1}{x} ; a=1, b=2$$

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