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In Exercises \(17-56,\) find the most general antiderivative or indefinite integral. You may need to try a solution and then adjust your guess. Check your answers by differentiation. $$\int(\sqrt{x}+\sqrt[3]{x}) d x$$

Short Answer

Expert verified
The most general antiderivative is \( \frac{2}{3}x^{3/2} + \frac{3}{4}x^{4/3} + C \).

Step by step solution

01

Identify the Components

The given integral is \( \int (\sqrt{x} + \sqrt[3]{x}) \, dx \). We need to find the antiderivative of each component separately: \( \sqrt{x} \) and \( \sqrt[3]{x} \).
02

Convert to Exponents

Rewrite the integral in terms of exponents: \( \sqrt{x} = x^{1/2} \) and \( \sqrt[3]{x} = x^{1/3} \). So, the integral becomes \( \int (x^{1/2} + x^{1/3}) \, dx \).
03

Find the Antiderivative of Each Component

Use the power rule for integration, \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \), where \( n eq -1 \). For \( x^{1/2} \):\[ \int x^{1/2} \, dx = \frac{x^{1/2 + 1}}{1/2 + 1} = \frac{x^{3/2}}{3/2} = \frac{2}{3}x^{3/2} \] For \( x^{1/3} \):\[ \int x^{1/3} \, dx = \frac{x^{1/3 + 1}}{1/3 + 1} = \frac{x^{4/3}}{4/3} = \frac{3}{4}x^{4/3} \]
04

Combine the Antiderivatives

The most general antiderivative is the sum of the two antiderivatives:\[ \frac{2}{3}x^{3/2} + \frac{3}{4}x^{4/3} + C \] where \( C \) is the constant of integration.
05

Verify by Differentiation

Differentiate the result \( \frac{2}{3}x^{3/2} + \frac{3}{4}x^{4/3} + C \) to ensure it matches the original integrand:\[ \frac{d}{dx} \left( \frac{2}{3}x^{3/2} \right) = \frac{2}{3} \cdot \frac{3}{2}x^{1/2} = x^{1/2} \]\[ \frac{d}{dx} \left( \frac{3}{4}x^{4/3} \right) = \frac{3}{4} \cdot \frac{4}{3}x^{1/3} = x^{1/3} \]Adding these gives \( x^{1/2} + x^{1/3} \), which is the original integrand.

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Key Concepts

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

Antiderivative
When you hear the term "antiderivative," it may sound a bit technical, but it's quite simple once you break it down. An antiderivative of a function is another function whose derivative brings us back to the original. That means, if you take the derivative of the antiderivative, you should recover the original function you started with. In the context of integrals, when we say find the indefinite integral of a function, we are looking for its antiderivative. This is because integrals can be seen as the reverse process of differentiation.
For example, if we have a function like \( f(x) = x^2 \), one of its antiderivatives is \( F(x) = \frac{1}{3}x^3 + C \). How do we know? By taking the derivative of \( F(x) \), which will give us back \( x^2 \). The constant \( C \) represents the constant of integration, which we'll discuss more later. Remember, different functions can have the same derivative, so the antiderivative is not just one function but a whole family of functions, which includes that mysterious constant \( C \).
This concept is vital as it forms the basis of solving indefinite integrals, which are integral signs without specific limits.
Power Rule for Integration
The power rule for integration is your best friend when dealing with functions raised to a power. It's a straightforward method to find antiderivatives. This rule is the integration equivalent to the power rule for differentiation. Here's how it works: when you have a function of the form \( x^n \), its antiderivative can be calculated as \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \), as long as \( n eq -1 \).
Let's dissect this rule a bit. You're essentially increasing the exponent by one and then dividing by that new exponent. When you apply this rule, it's easy to handle functions like \( x^{1/2} \) and \( x^{1/3} \), as showcased in the exercise solution. For \( x^{1/2} \), this turns into \( \frac{2}{3}x^{3/2} \), and for \( x^{1/3} \), it becomes \( \frac{3}{4}x^{4/3} \).
The beauty of the power rule is in its simplicity and power (pun intended). It allows us to quickly find antiderivatives for a wide class of functions without much fuss. Just always remember to add the constant of integration \( C \) because we're finding an indefinite integral.
Constant of Integration
The constant of integration, often denoted \( C \), might be small, but it carries a significant meaning in calculus. Every time you find an indefinite integral or antiderivative, this constant is appended to your result. Why does it matter? Well, when you differentiate a constant, you get zero. So, whether your function has a constant addition or not, its derivative remains unaffected.
This is why when you work with indefinite integrals, the result is not just one function, but a whole family of functions. The antiderivative of \( f(x) = x^2 \) isn't just \( \frac{1}{3}x^3 \). It's \( \frac{1}{3}x^3 + C \), where \( C \) can be any real number. That \( C \) accounts for all the possible vertical shifts of your antiderivative on a graph.
Neglecting the constant \( C \) is a common oversight that can lead to incorrect solutions, especially when evaluating expressions at certain points. Always include the constant of integration to get the "most general antiderivative" for your integrals. It's an integral (no pun intended) part of finding indefinite integrals!

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