Problem 41 - 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. $$\int \frac{t \sqrt{t}+\sqrt{t}}{t^{2}} d t$$

Step-by-Step Solution

Answer

The most general antiderivative is \( 2t^{1/2} - 2t^{-1/2} + C \).

Step 1: Simplify the Integrand

First, simplify the expression by dividing each term in the numerator by the denominator: \[ \int \frac{t \sqrt{t}}{t^2} + \frac{\sqrt{t}}{t^2} \, dt \]. Simplify each term: \( \frac{t \sqrt{t}}{t^2} = \frac{t^{3/2}}{t^2} = t^{-1/2} \) and \( \frac{\sqrt{t}}{t^2} = \frac{t^{1/2}}{t^2} = t^{-3/2} \). Thus, the integral becomes \( \int (t^{-1/2} + t^{-3/2}) \, dt \).

Step 2: Integrate Term by Term

Integrate each term separately. For \( \int t^{-1/2} \, dt \), use the power rule: Add 1 to the exponent and divide by the new exponent: \( \frac{t^{1/2}}{1/2} = 2t^{1/2} \).For \( \int t^{-3/2} \, dt \), again use the power rule: \( \frac{t^{-1/2}}{-1/2} = -2t^{-1/2} \).

Step 3: Write the General Antiderivative

Combine the results from the previous step to express the general antiderivative of the original function: \[ 2t^{1/2} - 2t^{-1/2} + C \], where \( C \) is the constant of integration.

Step 4: Verify by Differentiation

Differentiate the result \( 2t^{1/2} - 2t^{-1/2} + C \). The derivative of \( 2t^{1/2} \) is \( t^{-1/2} \), and the derivative of \( -2t^{-1/2} \) is \( -t^{-3/2} \). Thus, \( \frac{d}{dt} (2t^{1/2} - 2t^{-1/2} + C) = t^{-1/2} + t^{-3/2} \), which matches the original simplification in Step 1.

Key Concepts

Power Rule

The power rule is a straightforward technique used in calculus to find the derivative of a function of the form \( x^n \), where \( n \) is any real number. It is also applied inversely to find indefinite integrals, which is essential for solving various calculus problems.

• For derivatives, the power rule states that the derivative of \( x^n \) is \( n \, x^{n-1} \).
• For integration, the reverse process involves increasing the exponent by 1 and dividing by the new exponent. Thus, \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \) for \( n eq -1 \).

In the solution to the given integrand, the power rule was applied twice, once for each term. For \( t^{-1/2} \), the power rule gives us \( \frac{t^{1/2}}{1/2} = 2t^{1/2} \). Similarly, for \( t^{-3/2} \), it results in \( \frac{t^{-1/2}}{-1/2} = -2t^{-1/2} \).

As you can see, applying the power rule makes it easy to solve integrals that include powers of a variable. Just remember to increase the power, perform division, and adapt the rule based on whether it's differential or integration.

Antiderivative

An antiderivative of a function is another function whose derivative reproduces the original function. Finding an antiderivative is essentially reversing the process of differentiation.

• Antiderivatives can take many forms due to the constant of integration, which we'll discuss shortly.
• Finding an antiderivative is part of determining the indefinite integral of a function.

In the exercise, we are tasked with finding the most general antiderivative of the function \( \frac{t \sqrt{t} + \sqrt{t}}{t^2} \). This process involved simplifying the expression and then applying the integration power rule term by term. The result, \( 2t^{1/2} - 2t^{-1/2} \), is a function whose derivative is equal to the original function before integration.

The beauty of antiderivatives is that they provide us with a way to reconstruct a function from its rate of change. This makes them incredibly valuable in numerous applications of calculus across physics, engineering, and other fields.

Constant of Integration

In the context of indefinite integrals, the constant of integration, denoted as \( C \), is an essential component. It represents the fact that the derivative of a constant is zero, and hence there are infinitely many functions that have the same derivative.

• When you perform indefinite integration, every antiderivative of a function is accompanied by an arbitrary constant \( C \).
• This constant captures the family of functions that share the same derivative.

For the problem at hand, after finding the antiderivatives \( 2t^{1/2} \) and \( -2t^{-1/2} \), the general form of the solution is expressed as \( 2t^{1/2} - 2t^{-1/2} + C \).

Including \( C \) is crucial as it ensures the most general form of the solution is accounted for. Although this might appear as a simple addition, its significance in solving differential equations and understanding the bigger picture of integral curves cannot be overstated.