Problem 8
Question
Find the general antiderivative \(F(x)+C\) for each of the following. $$ f(x)=7 x^{-3 / 4} $$
Step-by-Step Solution
Verified Answer
The general antiderivative is \(F(x) = 28x^{\frac{1}{4}} + C\).
1Step 1: Identify the function
The function provided is \(f(x) = 7x^{-\frac{3}{4}}\). Our goal is to find the general antiderivative \(F(x) + C\), where \(C\) is a constant of integration.
2Step 2: Apply the Antiderivative Rule for Power Functions
The formula for finding the antiderivative of a power function \(x^n\) is \(\frac{x^{n+1}}{n+1}\), provided \(n eq -1\). We start by identifying \(n = -\frac{3}{4}\) for our function.
3Step 3: Find the Antiderivative
Using the formula from Step 2, the antiderivative is \(F(x) = \frac{7x^{-\frac{3}{4} + 1}}{-\frac{3}{4} + 1}\). Simplifying the exponent, we have \(-\frac{3}{4} + 1 = \frac{1}{4}\), so \(F(x) = \frac{7x^{\frac{1}{4}}}{\frac{1}{4}}\).
4Step 4: Simplify the Expression
To simplify \(\frac{7x^{\frac{1}{4}}}{\frac{1}{4}}\), note that dividing by \(\frac{1}{4}\) is the same as multiplying by 4. Thus, \(F(x) = 28x^{\frac{1}{4}}\).
5Step 5: Add Constant of Integration
The general form of the antiderivative is \(F(x) + C\), where \(C\) is the constant of integration. Therefore, the final answer is \(F(x) = 28x^{\frac{1}{4}} + C\).
Key Concepts
Power FunctionsConstant of IntegrationCalculusIntegration Techniques
Power Functions
Power functions are expressions of the form \( x^n \), where \( n \) is a real number called the exponent. They play a crucial role in calculus, especially when dealing with differentiation and integration.
When finding the antiderivative (or the indefinite integral) of a power function, the process involves adding one to the exponent and then dividing by the new exponent. This is known as the power rule for integration. It's important to remember that this rule cannot be used if the exponent \( n \) equals \(-1\), as it would lead to division by zero.
In the exercise we are looking at, the function was \( f(x) = 7x^{-\frac{3}{4}} \). We identify \( n = -\frac{3}{4} \). By following the power rule, we increased the exponent by 1, resulting in \( n + 1 = \frac{1}{4} \), and then divided the whole expression by the same \( n + 1 \).
When finding the antiderivative (or the indefinite integral) of a power function, the process involves adding one to the exponent and then dividing by the new exponent. This is known as the power rule for integration. It's important to remember that this rule cannot be used if the exponent \( n \) equals \(-1\), as it would lead to division by zero.
In the exercise we are looking at, the function was \( f(x) = 7x^{-\frac{3}{4}} \). We identify \( n = -\frac{3}{4} \). By following the power rule, we increased the exponent by 1, resulting in \( n + 1 = \frac{1}{4} \), and then divided the whole expression by the same \( n + 1 \).
Constant of Integration
The constant of integration \( C \) is a vital part of finding an antiderivative. When integrating a function, there are infinitely many functions that can result in the derivative or original function, which is why we add a constant \( C \).
This constant accounts for the fact that the derivative of a constant is zero. Therefore, the antiderivative must include it to cover all possible original functions. The exception to this occurs when boundary conditions or initial values are provided, leading to a particular solution rather than the general solution.
In the solution, after computing \( F(x) = 28x^{\frac{1}{4}} \), we add \( C \) to denote that there are many functions that may have given our original \( f(x) \) when differentiated.
This constant accounts for the fact that the derivative of a constant is zero. Therefore, the antiderivative must include it to cover all possible original functions. The exception to this occurs when boundary conditions or initial values are provided, leading to a particular solution rather than the general solution.
In the solution, after computing \( F(x) = 28x^{\frac{1}{4}} \), we add \( C \) to denote that there are many functions that may have given our original \( f(x) \) when differentiated.
Calculus
Calculus is a branch of mathematics focused on limits, functions, derivatives, integrals, and infinite series. It is used extensively in science and engineering for modeling dynamic relations among variables.
In particular, calculus allows for the manipulation of curves and the calculation of the area under these curves through integration. This is precisely what we do when finding an antiderivative. The antiderivative provides us with a function that, when differentiated, yields the curve represented by our initial function.
The process involves using established rules and techniques, such as the power rule, to rewrite complicated expressions as simplified areas or other dimensions of interest.
In particular, calculus allows for the manipulation of curves and the calculation of the area under these curves through integration. This is precisely what we do when finding an antiderivative. The antiderivative provides us with a function that, when differentiated, yields the curve represented by our initial function.
The process involves using established rules and techniques, such as the power rule, to rewrite complicated expressions as simplified areas or other dimensions of interest.
Integration Techniques
Integration techniques are methods used to find indefinite and definite integrals. They are essential for solving complex calculus problems. Some common techniques include substitution, integration by parts, and partial fraction decomposition.
For the exercise at hand, we used the power rule technique, which is directly applicable to power functions. This method is straightforward: modify the exponent and adjust the fraction accordingly. Mastery of this technique is key for efficiently solving and simplifying integrals with power functions.
More complex functions might require other techniques. Recognizing which technique to apply is crucial for successful calculus problem-solving.
For the exercise at hand, we used the power rule technique, which is directly applicable to power functions. This method is straightforward: modify the exponent and adjust the fraction accordingly. Mastery of this technique is key for efficiently solving and simplifying integrals with power functions.
More complex functions might require other techniques. Recognizing which technique to apply is crucial for successful calculus problem-solving.
Other exercises in this chapter
Problem 8
The population of the United States was \(3.9\) million in 1790 and 178 million in 1960 . If the rate of growth is assumed proportional to the number present, w
View solution Problem 8
First find the general solution (involving a constant \(\mathrm{C}\) ) for the given differential equation. Then find the particular solution that satisfies the
View solution Problem 8
Show that for a rectangle of given perimeter \(K\) the one with maximum area is a square.
View solution Problem 8
Identify the critical points. Then use (a) the First Derivative Test and (if possible) (b) the Second Derivative Test to decide which of the critical points giv
View solution