Problem 1 - Essential Calculus Early Transcendentals

Question

Find the most general antiderivative of the function. (Check your answer by differentiation.) $$f(x)=\frac{1}{2}+\frac{3}{4} x^{2}-\frac{4}{5} x^{3}$$

Step-by-Step Solution

Answer

The general antiderivative is \( F(x) = \frac{1}{2}x + \frac{1}{4}x^3 - \frac{1}{5}x^4 + C \).

Step 1: Identify the Power Rule for Antiderivatives

To find the antiderivative of a polynomial function, we apply the power rule for antiderivatives. For each term of the form \( ax^n \), the antiderivative is \( \frac{a}{n+1}x^{n+1} \).

Step 2: Apply the Power Rule to Each Term

For the term \( \frac{1}{2} \), treat it as \( \frac{1}{2}x^0 \), so the antiderivative is \( \frac{1}{2}x \). For \( \frac{3}{4}x^2 \), the antiderivative is \( \frac{3/4}{3}x^3 = \frac{1}{4}x^3 \). For \( -\frac{4}{5}x^3 \), the antiderivative is \( -\frac{4/5}{4}x^4 = -\frac{1}{5}x^4 \).

Step 3: Write the General Antiderivative

Combine all the individual antiderivatives, adding an arbitrary constant \( C \) to account for the general solution. The general antiderivative is:\[F(x) = \frac{1}{2}x + \frac{1}{4}x^3 - \frac{1}{5}x^4 + C\]

Step 4: Differentiate to Check the Solution

Differentiate the antiderivative \( F(x) = \frac{1}{2}x + \frac{1}{4}x^3 - \frac{1}{5}x^4 + C \). Using the power rule for differentiation, for each term:- The derivative of \( \frac{1}{2}x \) is \( \frac{1}{2}\).- The derivative of \( \frac{1}{4}x^3 \) is \( \frac{3}{4}x^2 \).- The derivative of \( -\frac{1}{5}x^4 \) is \( -\frac{4}{5}x^3 \). - The derivative of \( C \) is \( 0 \).The resulting function is \( f(x) = \frac{1}{2} + \frac{3}{4} x^2 - \frac{4}{5} x^3 \), which matches the original function.

Key Concepts

Power Rule for Antiderivatives

Understanding antiderivatives begins with the power rule, a fundamental tool in calculus. The power rule for antiderivatives states: for any term \( ax^n \), its antiderivative is given by \( \frac{a}{n+1}x^{n+1} + C \). This "+ C" is crucial as it represents the constant of integration, ensuring that we include all possible solutions.Let's break this down with a simple example:- If we have the term \( x^2 \), applying the power rule means finding the antiderivative by increasing the exponent by one (making it 3) and dividing by the new exponent. Therefore, the antiderivative is \( \frac{1}{3}x^3 \).The power rule offers a straightforward way to handle each term within a polynomial, making it easier to work with complex functions.This methodology is extensively useful in understanding how to construct antiderivatives systematically from derivatives.

Polynomial Antiderivatives

Polynomial functions, being sums of simple power terms, lend themselves naturally to the application of the power rule. When working with polynomial functions like \( f(x)=\frac{1}{2}+\frac{3}{4} x^{2}-\frac{4}{5} x^{3} \), each term can be tackled individually:- The constant \( \frac{1}{2} \) can be thought of as \( \frac{1}{2}x^0 \), whose antiderivative becomes \( \frac{1}{2}x \).- For \( \frac{3}{4}x^2 \), applying the power rule, the antiderivative becomes \( \frac{3/4}{3}x^3 = \frac{1}{4}x^3 \).- Similarly, for \( -\frac{4}{5}x^3 \), the antiderivative becomes \( -\frac{4/5}{4}x^4 = -\frac{1}{5}x^4 \).These steps combined give the general antiderivative:\[ F(x) = \frac{1}{2}x + \frac{1}{4}x^3 - \frac{1}{5}x^4 + C \]This approach ensures that each term in the polynomial is carefully addressed, leading to the correct general solution, inclusive of an arbitrary constant \( C \).

Differentiation Check

Verifying the correctness of an antiderivative involves differentiation. The process acts as a checkpoint to ensure our solution aligns with the original function.Starting with the antiderivative \( F(x) = \frac{1}{2}x + \frac{1}{4}x^3 - \frac{1}{5}x^4 + C \), differentiate each term:- The derivative of \( \frac{1}{2}x \) is simply \( \frac{1}{2} \).- For \( \frac{1}{4}x^3 \), it becomes \( \frac{3}{4}x^2 \) upon differentiation.- \( -\frac{1}{5}x^4 \) differentiates to \( -\frac{4}{5}x^3 \).- The derivative of the constant \( C \) is \( 0 \).Thus, the resulting function \( f(x) = \frac{1}{2} + \frac{3}{4} x^2 - \frac{4}{5} x^3 \) precisely matches the original function, confirming the antiderivative is correct. This step captures both the rigor and beauty of calculus: finding an antiderivative, yet ensuring its precision through differentiation.