Problem 2 - Calculus Single Variable

Question

Suppose that \(\int_{2}^{12} g(x) d x=-6\) and \(\int_{2}^{6} g(x) d x=-12\). Evaluate \(\int_{6}^{12} g(x) d x\).

Step-by-Step Solution

Answer

The value of \( \int_{6}^{12} g(x) \, dx \) is 6.

Step 1: Understand the Integral Splitting Property

The property of definite integrals that helps in solving this problem is \( \int_{a}^{b} f(x) \, dx = \int_{a}^{c} f(x) \, dx + \int_{c}^{b} f(x) \, dx \). This means that you can split an integral from any point \(c\) between \(a\) and \(b\).

Step 2: Apply Integral Splitting Property to the Given Integrals

We are given two integrals: \( \int_{2}^{12} g(x) \, dx = -6 \) and \( \int_{2}^{6} g(x) \, dx = -12 \). Applying the property: \[ \int_{2}^{12} g(x) \, dx = \int_{2}^{6} g(x) \, dx + \int_{6}^{12} g(x) \, dx \] Subtract the given integral \( \int_{2}^{6} g(x) \, dx = -12 \) from \( \int_{2}^{12} g(x) \, dx = -6 \).

Step 3: Calculate \( \int_{6}^{12} g(x) \, dx \)

Using the information and splitting property: \[ -6 = -12 + \int_{6}^{12} g(x) \, dx \] Add \(12\) to both sides to solve for \( \int_{6}^{12} g(x) \, dx \):\[ \int_{6}^{12} g(x) \, dx = -6 + 12 = 6 \] Hence, the value of \( \int_{6}^{12} g(x) \, dx \) is \(6\).

Key Concepts

Integral Splitting Property

The Integral Splitting Property is a fundamental concept in calculus that allows us to break down complex integrals into smaller, more manageable parts. This property is formally expressed as:

• \( \int_{a}^{b} f(x) \, dx = \int_{a}^{c} f(x) \, dx + \int_{c}^{b} f(x) \, dx \)

Here, you can pick any point \(c\) between \(a\) and \(b\) to split the integral. By doing this, you can solve complicated problems in pieces. It helps in integrating functions over multiple intervals by using known values of certain segments.
This splitting helps by providing flexibility and strategies to handle integrals, especially when you have results for some sub-intervals but need to find values for others.

Calculus

Calculus is a branch of mathematics centered around change. It consists of two main components: differentiation and integration. Integration, which is particularly relevant here, deals with accumulation and area under a curve.
The definite integral, such as \( \int_{a}^{b} g(x) \, dx \), computes the net area between the \(x\)-axis and the curve \(g(x)\) from \(x = a\) to \(x = b\). This is useful for evaluating mathematical properties that involve rates of change and accumulation.

• Differentiation focuses on instantaneous rates of change or slopes.
• Integration focuses on total accumulation over intervals.

Integrals allow us to aggregate values, which is key for understanding quantities like total distance, area, and volume in real-world applications.

Evaluating Integrals

Evaluating integrals involves finding the sum that accumulates the area under a curve over a specific interval.
To evaluate an integral like \( \int_{a}^{b} f(x) \, dx \), one needs a function \(f(x)\) and limits \(a\) and \(b\). When values for certain parts of an interval are known, such as \( \int_{a}^{c} f(x) \, dx \), splitting can aid in finding unknown sections as shown in the original exercise.In the given exercise:

• The integral \( \int_{2}^{12} g(x) \, dx \) is split using known values of \( \int_{2}^{6} g(x) \, dx \).
• This revealed the unknown portion \( \int_{6}^{12} g(x) \, dx \).

By using substitution and property applications, finding missing quantities becomes easier. This array of methods provides multiple pathways to solutions, enhancing problem-solving flexibility in calculus.