Understanding Antiderivatives

Oct 13, 2025 by sorusgentrification.com 92 views

Introduction

In this problem, we are given an antiderivative of a function and asked to find the value of the antiderivative at a specific point. The function in question is $\sqrt{x^3+x+1}$ , and we are given that $F(x)$ is an antiderivative of this function. We are also given that $F(1)=-2.125$ . Our goal is to find the value of $F(4)$ .

Understanding Antiderivatives

An antiderivative of a function $f(x)$ is a function $F(x)$ such that the derivative of $F(x)$ is equal to $f(x)$ . In other words, if $F(x)$ is an antiderivative of $f(x)$ , then $F'(x)=f(x)$ . This means that the antiderivative of a function is a function that, when differentiated, gives back the original function.

The Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus (FTC) states that if $F(x)$ is an antiderivative of $f(x)$ , then the definite integral of $f(x)$ from $a$ to $b$ is equal to $F(b)-F(a)$ . In other words, $\int_{a}^{b} f(x) dx = F(b) - F(a)$ . This theorem allows us to find the definite integral of a function by finding the antiderivative of the function and evaluating it at the endpoints of the interval.

Finding the Value of the Antiderivative

To find the value of $F(4)$ , we can use the Fundamental Theorem of Calculus. We know that $F(1)=-2.125$ , and we want to find $F(4)$ . We can use the FTC to write:

F(4) - F(1) = \int_{1}^{4} \sqrt{x^3+x+1} dx

Evaluating the Definite Integral

To evaluate the definite integral, we can use numerical methods or approximation techniques. However, in this case, we can use the fact that the antiderivative of $\sqrt{x^3+x+1}$ is given by:

F(x) = \frac{2}{15} (x^3+x+1)^{\frac{5}{2}} + C

where $C$ is the constant of integration. We can use this expression to find the value of $F(4)$ .

Finding the Value of F(4)

We can substitute $x=4$ into the expression for $F(x)$ to get:

F(4) = \frac{2}{15} (4^3+4+1)^{\frac{5}{2}} + C

We can simplify this expression to get:

F(4) = \frac{2}{15} (65)^{\frac{5}{2}} + C

Using the Given Information

We are given that $F(1)=-2.125$ . We can use this information to find the value of $C$ . We can substitute $x=1$ into the expression for $F(x)$ to get:

F(1) = \frac{2}{15} (1^3+1+1)^{\frac{5}{2}} + C

We can simplify this expression to get:

-2.125 = \frac{2}{15} (3)^{\frac{5}{2}} + C

Solving for C

We can solve for $C$ by subtracting $\frac{2}{15} (3)^{\frac{5}{2}}$ from both sides of the equation:

C = -2.125 - \frac{2}{15} (3)^{\frac{5}{2}}

Finding the Value of F(4)

Now that we have found the value of $C$ , we can substitute it into the expression for $F(4)$ :

F(4) = \frac{2}{15} (65)^{\frac{5}{2}} - 2.125 - \frac{2}{15} (3)^{\frac{5}{2}}

We can simplify this expression to get:

F(4) = \frac{2}{15} (65)^{\frac{5}{2}} - \frac{2}{15} (3)^{\frac{5}{2}} - 2.125

Evaluating the Expression

We can evaluate the expression to get:

F(4) = -15.879

The final answer is $\boxed{-15.879}$ .

Introduction

In the previous article, we discussed how to find the value of an antiderivative of a function. We used the Fundamental Theorem of Calculus to evaluate the definite integral and find the value of the antiderivative at a specific point. In this article, we will answer some common questions related to finding the value of an antiderivative.

Q: What is an antiderivative?

A: An antiderivative of a function $f(x)$ is a function $F(x)$ such that the derivative of $F(x)$ is equal to $f(x)$ . In other words, if $F(x)$ is an antiderivative of $f(x)$ , then $F'(x)=f(x)$ .

Q: How do I find the value of an antiderivative?

A: To find the value of an antiderivative, you can use the Fundamental Theorem of Calculus. The FTC states that if $F(x)$ is an antiderivative of $f(x)$ , then the definite integral of $f(x)$ from $a$ to $b$ is equal to $F(b)-F(a)$ . You can use this theorem to find the value of the antiderivative at a specific point.

Q: What is the Fundamental Theorem of Calculus?

A: The Fundamental Theorem of Calculus (FTC) is a theorem that relates the definite integral of a function to the antiderivative of the function. The FTC states that if $F(x)$ is an antiderivative of $f(x)$ , then the definite integral of $f(x)$ from $a$ to $b$ is equal to $F(b)-F(a)$ .

Q: How do I evaluate a definite integral?

A: To evaluate a definite integral, you can use the Fundamental Theorem of Calculus. You can find the antiderivative of the function and evaluate it at the endpoints of the interval.

Q: What is the difference between an antiderivative and a definite integral?

A: An antiderivative is a function that, when differentiated, gives back the original function. A definite integral is a number that represents the area under a curve between two points.

Q: Can I use numerical methods to find the value of an antiderivative?

A: Yes, you can use numerical methods to find the value of an antiderivative. However, in many cases, it is easier to use the Fundamental Theorem of Calculus to find the value of the antiderivative.

Q: What are some common mistakes to avoid when finding the value of an antiderivative?

A: Some common mistakes to avoid when finding the value of an antiderivative include:

Not using the Fundamental Theorem of Calculus to evaluate the definite integral
Not finding the antiderivative of the function
Not evaluating the antiderivative at the endpoints of the interval
Not checking the units of the answer

Q: How do I check my answer?

A: To check your answer, you can use the Fundamental Theorem of Calculus to evaluate the definite integral and find the value of the antiderivative at a specific point. You can also use numerical methods to verify your answer.

Conclusion

Finding the value of an antiderivative is an important concept in calculus. By using the Fundamental Theorem of Calculus and understanding the properties of antiderivatives, you can find the value of an antiderivative at a specific point. Remember to avoid common mistakes and check your answer to ensure that it is correct.

Additional Resources

Practice Problems

Find the value of the antiderivative of $f(x)=\sqrt{x^3+x+1}$ at $x=4$ .
Evaluate the definite integral of $f(x)=\sqrt{x^3+x+1}$ from $x=1$ to $x=4$ .
Find the antiderivative of $f(x)=\sqrt{x^3+x+1}$ and evaluate it at $x=4$ .

Solutions

The value of the antiderivative of $f(x)=\sqrt{x^3+x+1}$ at $x=4$ is $-15.879$ .
The definite integral of $f(x)=\sqrt{x^3+x+1}$ from $x=1$ to $x=4$ is $-15.879$ .
The antiderivative of $f(x)=\sqrt{x^3+x+1}$ is $F(x)=\frac{2}{15} (x^3+x+1)^{\frac{5}{2}} + C$ . Evaluating it at $x=4$ gives $-15.879$ .