Let $f$ Be A Continuous Function Such That $\int_{-4}^6 F(x) \, Dx = -6$ And $\int_1^{-4} F(x) \, Dx = -4$. What Is The Value Of $\int_1^6 \frac{1}{2} F(x) \, Dx$?

Let $f$ be a continuous function such that $\int_{-4}^6 f(x) \, dx = -6$ and $\int_1^{-4} f(x) \, dx = -4$. What is the value of $\int_1^6 \frac{1}{2} f(x) \, dx$?

In this article, we will explore the properties of definite integrals and how they can be used to solve problems involving continuous functions. We will use the given information about the function $f$ and its definite integrals to find the value of a new definite integral involving the function $f$.

One of the most important properties of definite integrals is the linearity property. This property states that the definite integral of a linear combination of functions is equal to the linear combination of their definite integrals. Mathematically, this can be expressed as:

$$\int_a^b (c_1 f(x) + c_2 g(x)) \, dx = c_1 \int_a^b f(x) \, dx + c_2 \int_a^b g(x) \, dx$$

where $c_1$ and $c_2$ are constants, and $f$ and $g$ are continuous functions.

Another important property of definite integrals is the additivity property. This property states that the definite integral of a function over a union of intervals is equal to the sum of its definite integrals over each interval. Mathematically, this can be expressed as:

$$\int_a^c f(x) \, dx = \int_a^b f(x) \, dx + \int_b^c f(x) \, dx$$

where $a$, $b$, and $c$ are constants, and $f$ is a continuous function.

We are given that the function $f$ is continuous and that the following definite integrals are equal to the given values:

$$\int_{-4}^6 f(x) \, dx = -6$$

$$\int_1^{-4} f(x) \, dx = -4$$

We are asked to find the value of the following definite integral:

$$\int_1^6 \frac{1}{2} f(x) \, dx$$

To solve this problem, we will use the linearity property of definite integrals. We can rewrite the given definite integral as:

$$\int_1^6 \frac{1}{2} f(x) \, dx = \frac{1}{2} \int_1^6 f(x) \, dx$$

We can then use the additivity property of definite integrals to rewrite the integral as:

$$\frac{1}{2} \int_1^6 f(x) \, dx = \frac{1}{2} \left( \int_1^{-4} f(x) \, dx + \int_{-4}^6 f(x) \, dx \right)$$

We are given that the following definite integrals are equal to the given values:

$$\int_1^{-4} f(x) \, dx = -4$$

$$\int_{-4}^6 f(x) \, dx = -6$$

We can substitute these values into the equation above to get:

$$\frac{1}{2} \left( \int_1^{-4} f(x) \, dx + \int_{-4}^6 f(x) \, dx \right) = \frac{1}{2} (-4 + (-6))$$

Simplifying the equation, we get:

$$\frac{1}{2} \left( \int_1^{-4} f(x) \, dx + \int_{-4}^6 f(x) \, dx \right) = \frac{1}{2} (-10)$$

$$\frac{1}{2} \left( \int_1^{-4} f(x) \, dx + \int_{-4}^6 f(x) \, dx \right) = -5$$

Therefore, the value of the definite integral is:

$$\int_1^6 \frac{1}{2} f(x) \, dx = -5$$

In this article, we used the properties of definite integrals to solve a problem involving a continuous function. We used the linearity property and the additivity property to rewrite the given definite integral and then substituted the given values to find the solution. The final answer is $\boxed{-5}$.

• [1] "Calculus" by Michael Spivak
• [2] "Calculus: Early Transcendentals" by James Stewart
• [3] "Introduction to Calculus" by Michael Sullivan

In our previous article, we explored the properties of definite integrals and how they can be used to solve problems involving continuous functions. We used the linearity property and the additivity property to find the value of a definite integral involving the function $f$. In this article, we will answer some common questions related to definite integrals and continuous functions.

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

A: A definite integral is a specific type of integral that has a fixed upper and lower limit. It is used to find the area under a curve between two specific points. An indefinite integral, on the other hand, is a general type of integral that does not have a fixed upper and lower limit. It is used to find the antiderivative of a function.

Q: How do I know if a function is continuous?

A: A function is continuous if it has no gaps or jumps in its graph. In other words, the function can be drawn without lifting the pencil from the paper. To check if a function is continuous, you can use the following criteria:

• The function is defined at every point in its domain.
• The function has a finite limit at every point in its domain.
• The function has a unique value at every point in its domain.

Q: What is the linearity property of definite integrals?

A: The linearity property of definite integrals states that the definite integral of a linear combination of functions is equal to the linear combination of their definite integrals. Mathematically, this can be expressed as:

$$\int_a^b (c_1 f(x) + c_2 g(x)) \, dx = c_1 \int_a^b f(x) \, dx + c_2 \int_a^b g(x) \, dx$$

where $c_1$ and $c_2$ are constants, and $f$ and $g$ are continuous functions.

Q: What is the additivity property of definite integrals?

A: The additivity property of definite integrals states that the definite integral of a function over a union of intervals is equal to the sum of its definite integrals over each interval. Mathematically, this can be expressed as:

$$\int_a^c f(x) \, dx = \int_a^b f(x) \, dx + \int_b^c f(x) \, dx$$

where $a$, $b$, and $c$ are constants, and $f$ is a continuous function.

Q: How do I use the linearity property and the additivity property to solve problems involving definite integrals?

A: To use the linearity property and the additivity property to solve problems involving definite integrals, you can follow these steps:

1. Identify the function and the intervals over which you want to find the definite integral.
2. Use the linearity property to rewrite the definite integral as a linear combination of simpler definite integrals.
3. Use the additivity property to rewrite the definite integral as the sum of its definite integrals over each interval.
4. Evaluate the definite integrals using the given information and the properties of definite integrals.

Q: What are some common mistakes to avoid when working with definite integrals?

A: Some common mistakes to avoid when working with definite integrals include:

• Failing to identify the function and the intervals over which you want to find the definite integral.
• Failing to use the linearity property and the additivity property to rewrite the definite integral.
• Failing to evaluate the definite integrals using the given information and the properties of definite integrals.
• Failing to check for continuity of the function.

In this article, we answered some common questions related to definite integrals and continuous functions. We discussed the linearity property and the additivity property of definite integrals, and we provided examples of how to use these properties to solve problems involving definite integrals. We also discussed some common mistakes to avoid when working with definite integrals. By following the steps outlined in this article, you can become proficient in working with definite integrals and continuous functions.