Chapter 11: Problem 10
Find the indefinite integral. $$\int \sqrt{2} d x$$
Short Answer
Expert verified
The indefinite integral of the function \(f(x) = \sqrt{2}\) is:
$$
\int \sqrt{2} dx = \sqrt{2}x + C
$$
Step by step solution
01
Write down the function to integrate
The function we're going to integrate is:
$$
f(x) = \sqrt{2}
$$
02
Integrate the function
Now we're going to find the indefinite integral of the function, i.e., its antiderivative. We'll use the power rule for integration, which states that
$$
\int x^n dx = \frac{x^{n+1}}{n+1} + C
$$
where n is a constant, and C is the constant of integration.
In our case, since the function is a constant function, we can rewrite the function as \(f(x) = x^0 \cdot \sqrt{2}\). Therefore, we can apply the power rule with \(n = 0\) to find the indefinite integral of the function.
03
Apply the power rule and find the antiderivative
Applying the power rule with \(n = 0\), we get:
$$
\int (\sqrt{2} x^0) dx = \frac{\sqrt{2} x^{0+1}}{0+1} + C = \sqrt{2}x + C
$$
04
Solve for the antiderivative
The antiderivative of the function \(f(x) = \sqrt{2}\) is:
$$
\int \sqrt{2} dx = \sqrt{2}x + C
$$
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Antiderivative
In calculus, finding an antiderivative means determining a function whose derivative gives back the original function. It's sometimes called the indefinite integral. When you calculate the indefinite integral of a function, you are essentially reversing differentiation. You’re looking for a function that, when differentiated, will return the initial function you started with.
For example, if you have a function like \(f(x) = \, \sqrt{2}\), the process involves finding some function \(F(x)\) such that the derivative \(F'(x)\) equals \(\sqrt{2}\). Here, \(F(x) = \, \sqrt{2}x + C\) works because, when you differentiate it, you get the original \(\sqrt{2}\) function. The constant \(C\) is a placeholder for any constant value because when differentiating, constants vanish.
For example, if you have a function like \(f(x) = \, \sqrt{2}\), the process involves finding some function \(F(x)\) such that the derivative \(F'(x)\) equals \(\sqrt{2}\). Here, \(F(x) = \, \sqrt{2}x + C\) works because, when you differentiate it, you get the original \(\sqrt{2}\) function. The constant \(C\) is a placeholder for any constant value because when differentiating, constants vanish.
- Antiderivatives are crucial for solving integrals.
- They help in finding the area under curves, among many other applications.
Power Rule
The power rule for integration is an essential tool for anyone learning calculus. It provides a straightforward method to find the antiderivative of functions that have the form of a variable raised to a power. The rule states:
\[ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \]where \(n\) is a real number, and \(C\) is the constant of integration.
This rule is versatile and applies to a wide range of functions. In the given exercise, we utilized the power rule for a constant function. Though \(\sqrt{2}\) is not explicitly expressed as \(x^n\), you can think of it as \(\sqrt{2}x^0\), with \(n = 0\). Applying the power rule with \(n = 0\) gives:
\[ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \]where \(n\) is a real number, and \(C\) is the constant of integration.
This rule is versatile and applies to a wide range of functions. In the given exercise, we utilized the power rule for a constant function. Though \(\sqrt{2}\) is not explicitly expressed as \(x^n\), you can think of it as \(\sqrt{2}x^0\), with \(n = 0\). Applying the power rule with \(n = 0\) gives:
- \( \int (\sqrt{2} x^0) \, dx = \frac{\sqrt{2} x^{0+1}}{0+1} + C \)
- It simplifies to \(\sqrt{2}x + C\).
Constant of Integration
When you solve an indefinite integral, you'll always encounter the constant of integration, denoted by \(C\). This constant represents an infinite set of answers rather than a single one. Why? Because in differentiation, any constant term disappears, meaning that there's an entire family of functions that could be the antiderivative.
For instance, if we integrate \(\sqrt{2}\), we arrive at \(\sqrt{2}x + C\). This \(C\) indicates that there could be multiple vertical shifts for the antiderivative function along the line.
For instance, if we integrate \(\sqrt{2}\), we arrive at \(\sqrt{2}x + C\). This \(C\) indicates that there could be multiple vertical shifts for the antiderivative function along the line.
- Different values of \(C\) correspond to different functions that are all antiderivatives of the same original function.
- Thus, \(C\) is a reminder that indefinite integrals describe a family of functions.
- In practical applications, additional information, like an initial condition, determines the specific value of \(C\).