Problem 37 - University Calculus: Early Transcendentals

Question

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

Step-by-Step Solution

Answer

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

Step 1: Simplify the Integral Expression

First, rewrite the integral in a form that is easier to integrate. The expression \(-\frac{2}{y^{1/4}}\) can be rewritten using negative exponents:\[\int \left(8y - 2y^{-1/4}\right) \, dy\].

Step 2: Integrate Each Term Separately

Integrate each term of the expression separately. Recall that the integral of \(y^n\) is \(\frac{y^{n+1}}{n+1} + C\):- For \(8y\), integrate to get \(8 \cdot \frac{y^{2}}{2} = 4y^2\).- For \(-2y^{-1/4}\), integrate to get \(-2 \cdot \frac{y^{3/4}}{3/4} = -\frac{8}{3}y^{3/4}\).

Step 3: Combine the Results

Combine the results from each term to write down the most general antiderivative:\[4y^2 - \frac{8}{3}y^{3/4} + C\] where \(C\) is the constant of integration.

Step 4: Differentiate to Check the Answer

Differentiate the result \(4y^2 - \frac{8}{3}y^{3/4} + C\) to verify:- Derivative of \(4y^2\) is \(8y\).- Derivative of \(-\frac{8}{3}y^{3/4}\) is \(-2y^{-1/4}\). Combine these results to get back the original function \(8y - 2y^{-1/4}\), confirming the solution is correct.

Key Concepts

Antiderivative

An antiderivative, also known as an indefinite integral, is a function whose derivative gives the original function. In simpler terms, if differentiating a function gives you a particular expression, then integrating that expression will bring you back to the original function plus a constant. This constant, denoted by \(C\), is essential because adding or subtracting a constant does not affect differentiation.
When finding an indefinite integral or an antiderivative, use the notation \(\int f(x) \, dx\). Here, the symbol \(\int\) represents the integration operation, while \(dx\) indicates the variable over which you are integrating. Finding the antiderivative helps in solving various mathematical problems, such as finding areas under curves or solving differential equations. Understanding how to work backward from a derivative to an antiderivative is key in calculus.

Integration Techniques

There are different techniques for integration, each suitable for different kinds of functions.

• **Basic Integration**: The most straightforward technique involves using a set of known integrals. For example, the integral of \(x^n\) is \(\frac{x^{n+1}}{n+1} + C\), as used in the solution.
• **Substitution Method**: Often useful when an integral is complicated, this method involves substituting part of the integrand (the function being integrated) with another variable to simplify the problem.
• **Integration by Parts**: This technique is applicable for integrands that are products of functions. It leverages the differentiation product rule in reverse.
• **Partial Fractions**: Suitable for rational functions, where you express the function as a sum of simpler fractions, which you then integrate individually.

In the solved exercise, basic integration was used efficiently by first simplifying the expression before integrating each term separately.

Negative Exponents

Negative exponents play an important role in integration and differentiation. They help in simplifying and rearranging algebraic expressions into forms that are easier to handle, especially in calculus.
Negative exponents are represented as \(x^{-n} = \frac{1}{x^n}\). This notation is particularly helpful in rewriting complicated fractions, transforming them into power expressions that are simpler to integrate or differentiate.
For example, in the exercise, the term \(-\frac{2}{y^{1/4}}\) was rewritten as \(-2y^{-1/4}\). Once written in this format, standard integration techniques apply, allowing you to integrate using the power rule: \(\int y^n \, dy = \frac{y^{n+1}}{n+1} + C\).
Mastering the use of negative exponents is crucial, as they frequently appear in calculus problems. Being comfortable with converting terms into and out of negative exponents will make solving such problems much simpler.