Problem 30 - Thomas Calculus in SI Units

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. $$\int\left(\frac{1}{7}-\frac{1}{y^{5 / 4}}\right) d y$$

Step-by-Step Solution

Answer

The antiderivative is \( \frac{1}{7}y - 4y^{-1/4} + C \).

Step 1: Understand the Integrand

The given integral is \( \int \left( \frac{1}{7} - \frac{1}{y^{5/4}} \right) dy \). This represents two separate functions: a constant \( \frac{1}{7} \) and a variable function \( -\frac{1}{y^{5/4}} \). We can use the linearity of integration to integrate each part separately.

Step 2: Integrate the Constant Term

Integrate the constant \( \frac{1}{7} \, dy \). The integral of a constant \( a \) is simply \( ay + C \), where \( C \) is a constant. Therefore, the integral of \( \frac{1}{7} \, dy \) is \( \frac{1}{7}y \).

Step 3: Rewrite the Variable Function

The function \( -\frac{1}{y^{5/4}} \) can be rewritten using exponent rules: \( -y^{-5/4} \). This form is easier to integrate than the initial fraction.

Step 4: Integrate the Variable Function

For the integral \( \int -y^{-5/4} \, dy \), use the power rule of integration, which is \( \int y^n \, dy = \frac{y^{n+1}}{n+1} + C \). Here \( n = -\frac{5}{4} \), so the integral is \( \frac{y^{-1/4}}{-1/4} = -4y^{-1/4} \).

Step 5: Combine the Results

Combine the integrals of both terms: \( \int \left( \frac{1}{7} - \frac{1}{y^{5/4}} \right) dy = \frac{1}{7}y - 4y^{-1/4} + C \), where \( C \) represents the constant of integration.

Step 6: Check by Differentiation

Differentiate \( \frac{1}{7}y - 4y^{-1/4} + C \) with respect to \( y \). The derivative is \( \frac{1}{7} - (-1/4 \cdot -4)y^{-5/4} = \frac{1}{7} - \frac{1}{y^{5/4}} \), which matches the original integrand.

Key Concepts

Indefinite Integral

The indefininte integral, also known as the antiderivative, is a fundamental concept in calculus. Think of it as the reverse process of differentiation. When you have a function and you want to find its integral, you're basically looking for another function whose derivative gives you the original function. In our exercise, the indefinite integral is written as:\[ \int \left(\frac{1}{7} - \frac{1}{y^{5/4}}\right) \, dy \]The solution to this integral will be a function plus a constant \( C \), since differentiation of a constant is zero. This constant appears because when taking antiderivatives, any constant can be there without affecting the derivative. Finding the indefinite integral involves techniques like recognizing parts of the function that can be integrated separately. This is where concepts like the power rule and linearity come into play, allowing you to solve it term by term.

Power Rule of Integration

The power rule of integration is a handy tool for dealing with polynomials, and it's essential for solving integrals involving variable exponents like in our example. The rule states that for any real number \( n eq -1 \):\[ \int y^n \, dy = \frac{y^{n+1}}{n+1} + C \]This rule is easy to apply when the function is expressed in the form \( y^n \). For instance, in our exercise, the expression \( -y^{-5/4} \) needs to be transformed using the power rule.Applying the rule here, when you integrate \( -y^{-5/4} \), you increase the exponent by 1, making it \( -1/4 \), and then divide by the new exponent to obtain the term:\[ -4y^{-1/4} \]Always remember to add \( C \) to represent the constant of integration. Once you find the antiderivative using the power rule, don’t forget to differentiate it again to check your work!

Linearity of Integration

Linearity of integration is a principle that makes solving integrals easier, especially when you have a function that can be broken down into simpler parts. It states that the integral of a sum of functions is equal to the sum of the integrals of each function, allowing you to treat each term independently. Mathematically, this is expressed as:\[ \int (f(y) + g(y)) \, dy = \int f(y) \, dy + \int g(y) \, dy \]In this exercise, the function \( \frac{1}{7} - \frac{1}{y^{5/4}} \) can be divided into two parts: a constant term and a term with a variable exponent. The linearity allows us to integrate the constant term \( \frac{1}{7} \) separately as a straightforward calculation:\[ \frac{1}{7}y \]While the variable term \( -y^{-5/4} \) requires applying the power rule. By using linearity, integration becomes more manageable and straightforward as you can tackle one piece at a time, making the overall process less daunting.