Problem 29 - 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. $$\int\left(8 y-\frac{2}{y^{1 / 4}}\right) d y$$

Step-by-Step Solution

Answer

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

Step 1: Break Down the Integral

We are given the integral \( \int\left(8y - \frac{2}{y^{1/4}}\right) dy\). This is a combination of two simpler integrals: \( \int 8y \, dy \) and \( \int -\frac{2}{y^{1/4}} \, dy \). We'll solve each part separately.

Step 2: Integrate \( 8y \)

First, integrate \( 8y \, dy \). We recognize this as a standard power rule integration problem. The integral of \( y^n \) with respect to \( y \) is \( \frac{y^{n+1}}{n+1} \). Here, \( n = 1 \), so:\[\int 8y \, dy = 8 \cdot \frac{y^{1+1}}{1+1} = 8 \cdot \frac{y^2}{2} = 4y^2\]

Step 3: Integrate \( -\frac{2}{y^{1/4}} \)

Next, convert the integrand \( -\frac{2}{y^{1/4}} \) to a power of \( y \):\( -\frac{2}{y^{1/4}} = -2y^{-1/4} \).Apply the power rule for integration: the integral of \( y^n \) is \( \frac{y^{n+1}}{n+1} \), with \( n = -\frac{1}{4} \):\[\int -2y^{-1/4} \, dy = -2 \cdot \frac{y^{-1/4+1}}{-1/4+1} = -2 \cdot \frac{y^{3/4}}{3/4} = -\frac{8}{3}y^{3/4}\]

Step 4: Combine Results and Add Constant of Integration

The antiderivative is the combination of the results from Steps 2 and 3, plus the constant of integration \( C \):\[\int\left(8y - \frac{2}{y^{1/4}}\right) dy = 4y^2 - \frac{8}{3}y^{3/4} + C\]

Step 5: Verify by Differentiation

Differentiate \( 4y^2 - \frac{8}{3}y^{3/4} + C \). The derivative of \( 4y^2 \) is \( 8y \) and the derivative of \(-\frac{8}{3}y^{3/4}\) uses the power rule: \(-\frac{8}{3} \cdot \frac{3}{4} y^{-1/4} = -2y^{-1/4} \). Constant \( C \) differentiates to 0.Bringing it all together, we have \( 8y - \frac{2}{y^{1/4}} \), which matches the original integrand.

Key Concepts

Understanding the Power Rule for Integration

When tackling integrals, the power rule is a fundamental concept worth mastering. It's an essential technique, especially for simple polynomial terms. The power rule states that the integral of \( y^n \) with respect to \( y \) is \( \frac{y^{n+1}}{n+1} \), provided \( n eq -1 \). This rule allows us to increase the exponent by 1 and then divide by the new exponent.

• For example, consider \( \int 8y \, dy \). Recognize \( y \) here has an exponent of 1 (since \( y = y^1 \)).
• Following the power rule, integrate it as \( 8 \cdot \frac{y^{1+1}}{1+1} = 4y^2 \).

It's essential to apply the power rule correctly, ensuring each new expression adheres to this simple format. This clarity simplifies the process and allows students to independently solve similar problems.

Exploring the Concept of Antiderivatives

Antiderivatives, often referred to as indefinite integrals, are a pivotal part of calculus. They are the reverse process of differentiation. Finding an antiderivative means identifying a function whose derivative equals the given function. When we integrate \( 8y - \frac{2}{y^{1/4}} \), we're essentially searching for its antiderivative.

• Its result, \( 4y^2 - \frac{8}{3}y^{3/4} + C \), where \( C \) is the constant of integration.
• This constant \( C \) reflects the notion that there are infinitely many possible antiderivatives for a given function, differing only by a constant.

Thus, the task of finding antiderivatives is about reversing differentiation, yielding broader families of functions related by constants.

Integration by Parts: A More Advanced Tool

Though not used directly for the given problem, integration by parts is a powerful technique beneficial in many integration challenges involving products of functions. This method stems from the product rule of differentiation and can be formulated as:\[\int u \, dv = uv - \int v \, du\]where \( u \) and \( dv \) are parts of the original integral.

• Choose \( u \) such that its derivative \( du \) is simpler, and select \( dv \) such that its integral \( v \) is easily determined.
• This technique is particularly useful when handling integrals of products involving polynomials and exponentials, or logs.

Understanding when and how to apply integration by parts requires practice, but mastering this opens up possibilities for handling more complex integrations.