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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. $$\int\left(8 y-\frac{2}{y^{1 / 4}}\right) d y$$

Short Answer

Expert verified
The antiderivative is \( 4y^2 - \frac{8}{3}y^{3/4} + C \).

Step by step solution

01

Break Down the Integral

The integral to solve is \( \int \left(8y - \frac{2}{y^{1/4}}\right) dy \). We’ll handle each term separately. The integral is the sum of two simpler integrals: \( \int 8y \, dy \) and \( \int -\frac{2}{y^{1/4}} \, dy \).
02

Integrate the First Term

For the first term \( \int 8y \, dy \), use the power rule for integration: \( \int y^n \, dy = \frac{y^{n+1}}{n+1} + C \). Here, \( n = 1 \), so \( \int 8y \, dy = 8 \cdot \frac{y^{2}}{2} = 4y^2 \).
03

Integrate the Second Term

For the second term \( \int -\frac{2}{y^{1/4}} \, dy \), rewrite it using exponents: \( \int -2y^{-1/4} \, dy \). Again, apply the power rule: \( \int y^n \, dy = \frac{y^{n+1}}{n+1} + C \). Here, \( n = -\frac{1}{4} \), so the integral is \( \frac{-2y^{3/4}}{3/4} = -\frac{8}{3}y^{3/4} \).
04

Combine Results and Add Constant

Combine the results from Steps 2 and 3 to form the most general antiderivative: \( 4y^2 - \frac{8}{3}y^{3/4} + C \), where \( C \) is the constant of integration.
05

Differentiate to Check

Differentiate \( 4y^2 - \frac{8}{3}y^{3/4} + C \) to verify correctness: \( \frac{d}{dy}(4y^2) = 8y \) and \( \frac{d}{dy}(-\frac{8}{3}y^{3/4}) = -2y^{-1/4} \). Combining these gives \( 8y - \frac{2}{y^{1/4}} \), which matches the original integrand.

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Key Concepts

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

Indefinite Integral
An indefinite integral is a core concept in calculus, often referred to as the antiderivative of a function. It represents the process of finding a function whose derivative is the given function. In other words, if you have a derivative and want to go back to the original function, you use indefinite integration. When we write an indefinite integral, it is typically denoted by the integral symbol \( \int \), followed by the function and the differential (such as \( dy \)). The result includes a constant \( C \), known as the constant of integration.

Why include a constant? Because during differentiation, constants disappear. So when you integrate, any constant could have been in the original function.
  • Integral Sign: \( \int \)
  • Function being integrated: The expression after the integral sign
  • Differential: Indicates the variable of integration (e.g., \( dy \) implies we integrate with respect to \( y \))
  • Constant of Integration: \( C \), representing any constant value
Understanding indefinite integrals is essential because they allow us to recreate a broader family of functions from derivatives. Remember, solving indefinite integrals is akin to reversing the process of differentiation.
Integration Techniques
Integration techniques are a set of methods or strategies used to find integrals of functions. These techniques are essential, especially when dealing with complex expressions, allowing us to simplify or manipulate an expression into an integrable form. There are several techniques, each with its unique utility. Let's look at some primary methods:
  • Substitution: Used when an integral involves a function and its derivative. You make a substitution to simplify the integral.
  • Integration by Parts: Useful for products of functions, based on the product rule for differentiation.
  • Partial Fractions: Applies to rational functions, separating them into simpler fractions to integrate each part easily.
  • Trigonometric Substitution: Deals with integrals involving \( \sqrt{a^2 - x^2} \) or similar forms.
  • Specific Functions: Apply standard rules for exponential, logarithmic, and trigonometric functions.
In our example, we utilized one of the basic integration techniques by breaking down the integral into simpler, more manageable terms. By handling each term individually, using methods like rewriting exponents, solving the integral becomes easier and much more systematic.
Power Rule for Integration
The power rule for integration is a straightforward yet powerful technique used to integrate functions involving powers of \( x \). This rule is similar to its differential counterpart but works in reverse:\[\int x^n \, dx = \frac{x^{n+1}}{n+1} + C \]Here, \( n \) represents the exponent. The power rule won't work for \( n = -1 \), as this forms a natural logarithm instead (i.e., \( \int \frac{1}{x} \, dx = \ln |x| + C \)).

To see the power rule in action, let's break it down using the steps outlined in our exercise.
  • For the term \( 8y \), \( n = 1 \). Applying the power rule, we get \( 4y^2 \), after inserting the constant multiple \( 8 \).

  • For \( -\frac{2}{y^{1/4}} \), rewrite as \( -2y^{-1/4} \). Using the power rule again for \( n = -1/4 \), results in \( -\frac{8}{3}y^{3/4} \).
This rule provides a quick route to integrate polynomial and polynomial-type expressions, making it an invaluable tool in calculus. By practicing and applying the power rule regularly, you can tackle many integration problems with confidence and precision.

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