Problem 59 - Calculus Early Transcendentals

Question

For the following functions \(f\), find the antiderivative \(F\) that satisfies the given condition. $$f(x)=x^{5}-2 x^{-2}+1 ; F(1)=0$$

Step-by-Step Solution

Answer

Question: Find the antiderivative of the function \(f(x) = x^5 - 2x^{-2} + 1\) that satisfies the condition \(F(1) = 0\). Answer: The antiderivative of the function is \(F(x) = \frac{x^6}{6} + 2x + x - \frac{19}{6}\).

Step 1: Find the general antiderivative of \(f(x)\)

To find the antiderivative of \(f(x)\), simply integrate the function term by term. We are given: $$f(x)=x^{5}-2 x^{-2}+1$$ Compute the antiderivative of each component: $$\int x^5 dx = \int x^n dx = \frac{x^{n+1}}{n+1} = \frac{x^6}{6}$$ $$\int -2x^{-2} dx = \int -2x^n dx = -(2)\frac{x^{n+1}}{n+1} = -2\frac{x^{-1}}{-1} = 2x$$ $$\int 1 dx = x$$ Now, combine these integrals to obtain the general antiderivative \(F(x)\): $$F(x) = \frac{x^6}{6} + 2x + x + C$$

Step 2: Use the given condition to find the constant of integration C

We are given the condition that \(F(1) = 0\). Plug the value of \(x=1\) into the general antiderivative to solve for the constant \(C\): $$0 = \frac{1^6}{6} + 2(1) + 1 + C$$ Now, solve for \(C\): $$0 = \frac{1}{6} + 2 + 1 + C$$ $$C = -\frac{1}{6} - 3 = -\frac{19}{6}$$

Step 3: Write the final antiderivative with the specific constant

Now that we have found the value of the constant \(C\), we can write the final antiderivative \(F(x)\): $$F(x) = \frac{x^6}{6} + 2x + x - \frac{19}{6}$$

Key Concepts

Integration

Integration is a fundamental concept in calculus, often described as the reverse process of differentiation. It helps us find the antiderivative of a function, giving us a new function whose derivative is the original function we started with. In practical terms, integration allows us to calculate the area under a curve on a graph.
When integrating polynomial functions like \(f(x) = x^5 - 2x^{-2} + 1\), we deal with each term separately. The power rule for integration aids in this process. According to the power rule, to integrate \(x^n\), where \(n\) is any real number except \(-1\), the result is \(\frac{x^{n+1}}{n+1}\).

• For \(x^5\), applying the power rule yields \(\frac{x^6}{6}\).
• For \(-2x^{-2}\), transforming it to \(-2\cdot\frac{x^{-1}}{-1}\) gives \(2x\).
• For the constant \(1\), integration yields \(x\), since integrating a constant returns a linear function.

Once all terms are integrated, they are summed to form the antiderivative. Remember, integration of a function represents the accumulation of values, where each term contributes to the total result.

Constant of Integration

Upon integrating a function, the result is not entirely complete without considering the constant of integration, often denoted by \(C\). This constant represents an entire family of functions that are antiderivatives of the original function, differing only by a constant. The reason for this is that when differentiating, constant terms disappear, leaving their original values indeterminate when we reverse the process.
A specific value of \(C\) can be determined if additional information, such as initial conditions, is given. In the exercise provided, we have the condition \(F(1) = 0\). By plugging \(x = 1\) into the antiderivative \(F(x) = \frac{x^6}{6} + 2x + x + C\), we can solve for \(C\). This allows us to find a unique antiderivative that fits the condition:

• Set up the equation: \(0 = \frac{1}{6} + 2 + 1 + C\)
• Simplify: \(0 = \frac{3}{6} + C\)
• Solve for \(C\): \(C = -\frac{19}{6}\)

With \(C\) defined, the particular solution to the integral not only represents a solution but fits the specific scenario outlined by the condition.

Polynomial Functions

Polynomial functions are expressions composed of variables raised to various powers and multiplied by coefficients, such as \(x^5 - 2x^{-2} + 1\) in the exercise. They are named for their variable power, which characterizes each term within the function, creating a varied landscape of behavior as \(x\) changes.
When dealing with polynomials, integration aids us in finding the antiderivative by tackling each term independently. This is straightforward with polynomial terms since rules like the power rule make integration a methodical process. Here's how the integration of individual polynomial terms simplified our task:

• The term \(x^5\) integrates to \(\frac{x^6}{6}\).
• The term \(-2x^{-2}\) when integrated returned \(2x\), after applying the power rule correctly.
• The constant \(1\) integrates to \(x\).

Once each term is integrated, they can be summed to form the polynomial antiderivative. Polynomial functions are particularly versatile because they can model a variety of real-world situations, from simple linear models to more complex behaviors as described by their higher-order terms.