Problem 34 - 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. $$\int \frac{4+\sqrt{t}}{t^{3}} d t$$

Step-by-Step Solution

Answer

The antiderivative is \(-2t^{-2} - \frac{2}{5}t^{-\frac{5}{2}} + C\).

Step 1: Simplify the Integrand

First, break down the fraction so that it becomes easier to integrate each term separately. The expression inside the integral is \( \frac{4}{t^3} + \frac{\sqrt{t}}{t^3} \). Simplify further: \( \frac{4}{t^3} \) is already in a usable form, and \( \frac{\sqrt{t}}{t^3} = t^{-\frac{1}{2} - 3} = t^{-\frac{7}{2}} \). Now the integral becomes: \[ \int \left( 4t^{-3} + t^{-\frac{7}{2}} \right) dt \].

Step 2: Integrate Each Term Separately

Apply the power rule for integration to each term: The power rule states that \( \int t^n dt = \frac{t^{n+1}}{n+1} + C \), given \( n eq -1 \). Integrate \(4t^{-3}\) to get \(-2t^{-2}\), and integrate \(t^{-\frac{7}{2}}\) to get \(-\frac{2}{5}t^{-\frac{5}{2}}\). Therefore, the integral becomes \(-2t^{-2} - \frac{2}{5}t^{-\frac{5}{2}} + C\).

Step 3: Verify by Differentiation

Differentiate the result to verify it's the correct antiderivative. Start by differentiating \(-2t^{-2}\) which gives \(4t^{-3}\), and \(-\frac{2}{5}t^{-\frac{5}{2}}\) which gives \( t^{-\frac{7}{2}} \). Therefore, the derivative of our solution is \(4t^{-3} + t^{-\frac{7}{2}} \), which matches our integrand, confirming the solution is correct.

Key Concepts

Antiderivative

In calculus, an antiderivative is a function whose derivative matches the given function. Simply put, if you differentiate an antiderivative, you should get back the original function.
This is why sometimes antiderivatives are also called indefinite integrals.

• Finding an antiderivative involves reversing the process of differentiation.
• While differentiation gives us the rate of change of a function, integration helps us find the original function from the rate.

When you integrate a function, you look for the most general form of the original function that could have been differentiated to provide the integrand. This involves adding a constant, usually referred to as "+ C," because the derivative of any constant is zero, so the constant's presence will not affect the derivative result.
It's crucial to check your work by differentiating your result to see if you return to the original function. This step confirms that you have found the correct antiderivative.

Integration Techniques

To handle integrals like the one in our exercise, we often rely on various integration techniques. These techniques allow us to simplify complex problems into manageable parts.

• The goal is to transform complicated expressions into simpler ones that are easier to integrate.
• The most common technique involves rewriting or simplifying the integrand.

For instance, in our original problem, the integration starts by breaking down the fraction and rewriting it, which makes it easier to apply basic rules.
Simplification often involves algebraic manipulation, like breaking down a fraction or separating terms within the integrand.
This simplification prepares the ground for using basic integration rules effectively, such as the power rule, which is among the most frequently used methods in introductory calculus.

Power Rule for Integration

The power rule is one of the simplest yet powerful techniques in calculus for finding antiderivatives. This rule is particularly useful when dealing with power functions.

• The power rule states: \( \int t^n \ dt = \frac{t^{n+1}}{n+1} + C \), as long as \( n eq -1 \).
• It is directly derived from understanding how powers work when differentiating and integrating.

In the exercise, we applied the power rule to each term of our integrand separately.
By simplifying the integrand terms, we had expressions like \( t^{-3} \) and \( t^{-\frac{7}{2}} \), which fit nicely with the power rule. You first increase the exponent by one and then divide the term by the new exponent.
After integrating using this method, don't forget to add the constant \( + C \) to account for all possible antiderivatives. This approach quickly turns complex-looking problems into straightforward solutions if you understand how to correctly apply the power rule.