Problem 1 - Thomas Calculus

Question

In Exercises \(1-16,\) find an antiderivative for each function. Do as many as you can mentally. Check your answers by differentiation. a. 2\(x \quad\) b. \(x^{2} \quad\) c. \(x^{2}-2 x+1\)

Step-by-Step Solution

Answer

a. \(x^2 + C\); b. \(\frac{x^3}{3} + C\); c. \(\frac{x^3}{3} - x^2 + x + C\).

Step 1: Antiderivative of 2x

The antiderivative of a function is found by reversing the differentiation process. For the function \(2x\), an antiderivative can be found by applying the power rule in reverse. Start with \(f(x) = 2x\), and increase the exponent by 1 to get \(x^{2}\). Then divide by the new exponent:\[F(x) = \int 2x \, dx = x^2 + C\]where \(C\) is the constant of integration.

Step 2: Verify Antiderivative of 2x by Differentiation

Differentiate \(F(x) = x^2 + C\) to verify:\[\frac{d}{dx}(x^2 + C) = 2x\]This confirms that the antiderivative is correct because it matches the original function.

Step 3: Antiderivative of x^2

Apply the reverse power rule to \(x^2\). Increase the exponent by 1 to get \(x^3\) and divide by the new exponent:\[F(x) = \int x^2 \, dx = \frac{x^3}{3} + C\]where \(C\) is the constant of integration.

Step 4: Verify Antiderivative of x^2 by Differentiation

Differentiate \(F(x) = \frac{x^3}{3} + C\):\[\frac{d}{dx}\left(\frac{x^3}{3} + C\right) = x^2\]This confirms that the antiderivative is correct because it matches the original function.

Step 5: Antiderivative of x^2 - 2x + 1

Apply the reverse power rule to each term separately:- Antiderivative of \(x^2\): \(\frac{x^3}{3}\)- Antiderivative of \(-2x\): \(-x^2\)- Antiderivative of \(1\): \(x\)Combine these to form the antiderivative:\[F(x) = \int (x^2 - 2x + 1) \, dx = \frac{x^3}{3} - x^2 + x + C\]where \(C\) is the constant of integration.

Step 6: Verify Antiderivative of x^2 - 2x + 1 by Differentiation

Differentiate \(F(x) = \frac{x^3}{3} - x^2 + x + C\):\[\frac{d}{dx}\left(\frac{x^3}{3} - x^2 + x + C\right) = x^2 - 2x + 1\]This confirms that the antiderivative is correct because it matches the original function.

Key Concepts

Power Rule

The power rule is a foundational concept in calculus, used primarily to work with polynomials. It simplifies the process of finding derivatives and antiderivatives. In differentiation, the power rule states:

• The derivative of \(x^n\) is \(nx^{n-1}\).
• For example, the derivative of \(x^2\) is \(2x\).

When reversed, this power rule assists in finding antiderivatives.

• To find the antiderivative of \(x^n\), increase the exponent by 1 to get \(x^{n+1}\).
• Then, divide by the new exponent, resulting in \(\frac{x^{n+1}}{n+1}\).

This is how we reverse the process from differentiation to find the integral, which is another term for antiderivative in this context. This rule is simple, yet powerful, allowing us to solve a broad array of problems involving polynomial functions.

Differentiation

Differentiation is the process of finding a function's derivative, which represents its rate of change at any given point.

• It transforms a function into its derivative, providing insight into the function's behavior.
• For instance, differentiating \(x^2 + C\) yields \(2x\), confirming that \(x^2\) is the antiderivative of \(2x\).

This ability to check whether a solution to an antiderivative problem is correct is crucial.

• If differentiation of the antiderivative takes you back to the original function, the solution is verified.

Differentiation is, therefore, both a tool for analysis and a method for verification in calculus.

Integration

Integration, often symbolized by the integral sign \(\int\), is the reverse process of differentiation. It helps to find antiderivatives, which are also known as integral functions.

• The process of integration involves accumulating the quantity represented by a function.
• For instance, integrating \(2x\) results in \(x^2 + C\).

Through integration, you are essentially adding up an infinite number of infinitesimally small quantities, providing a total value or accumulated change.

• Thus, integration is fundamental for calculating areas, volumes, and the antiderivatives of functions.

Understanding integration prepares you to solve complex real-world problems involving continuous change.

Constant of Integration

When performing integration to find an antiderivative, the constant of integration \(C\) is added to account for any constant term that could have been lost during differentiation.

• The constant \(C\) reflects the vertical shift of antiderivatives along the y-axis.
• Since differentiating a constant results in zero, the specific constant value present in the original function isn't known, necessitating the addition of \(C\).

For any function, there are infinitely many antiderivatives, collectively described by the addition of this constant.

• For instance, if \(F(x) = x^2\) is an antiderivative of \(2x\), then \(F(x) = x^2 + C\) is the general form of the antiderivative.

This makes the constant of integration an essential part of solving integration problems, maintaining accuracy alongside flexibility. Understanding \(C\) strengthens your grasp of antiderivatives and ensures precision in calculus work.