Problem 13 - Calculus

Question

Suppose that \(\int_{0}^{1} f(x) d x=2, \int_{1}^{2} f(x) d x=3, \int_{0}^{1} g(x) d x=-1\)and \(\int_{0}^{2} g(x) d x=4\). Use properties of definite integrals (linearity, interval additivity, and so on) to calculate each of the integrals. $$ \int_{2}^{1}[2 f(s)+5 g(s)] d s $$

Step-by-Step Solution

Answer

The value of the integral is \(-31\).

Step 1: Flip the Integration Limits

The integral is given as \( \int_{2}^{1}[2 f(s)+5 g(s)] d s \). We can change the limits of integration from \(2\) to \(1\) into \(1\) to \(2\) by changing the sign: \[ \int_{2}^{1}[2 f(s)+5 g(s)] d s = -\int_{1}^{2}[2 f(s)+5 g(s)] d s. \]

Step 2: Apply Linearity of Integrals

Using the linearity property of integrals, we can separate the integral into two parts: \[ -\int_{1}^{2}[2 f(s) + 5 g(s)] ds = -\left( \int_{1}^{2} 2 f(s) ds + \int_{1}^{2} 5 g(s) ds \right). \] This simplifies further by factoring the constants out of the integrals to: \[ -\left( 2 \int_{1}^{2} f(s) ds + 5 \int_{1}^{2} g(s) ds \right). \]

Step 3: Evaluate \(\int_{1}^{2} f(s) ds\)

From the problem statement, we know that \(\int_{1}^{2} f(x) dx = 3\). Thus, \[ 2 \int_{1}^{2} f(s) ds = 2 \times 3 = 6. \]

Step 4: Evaluate \(\int_{1}^{2} g(s) ds\)

To find \(\int_{1}^{2} g(s) ds\), we can use interval additivity: We have, \[ \int_{0}^{2} g(s) ds = \int_{0}^{1} g(s) ds + \int_{1}^{2} g(s) ds. \] Given \(\int_{0}^{2} g(x) dx = 4\) and \(\int_{0}^{1} g(x) dx = -1\), solve for \(\int_{1}^{2} g(x) dx\): \[ 4 = -1 + \int_{1}^{2} g(s) ds \quad \Rightarrow \quad \int_{1}^{2} g(s) ds = 4 + 1 = 5. \]

Step 5: Substitute and Compute

Substitute the values from Steps 3 and 4 into the equation from Step 2: \[ -\left(2 \times 3 + 5 \times 5\right) = -(6 + 25) = -31. \]

Key Concepts

Linearity of Integrals

The linearity of integrals is a fundamental property that simplifies complex integral calculations significantly. This concept allows us to break down an integral that involves a sum of functions into separate integrals. This can make the overall process much easier to handle and understand.

In the exercise, we used the linearity property to separate the integral of a linear combination of functions, \( [2f(s)+ 5g(s)] \), into individual parts. This separation can be expressed as:

• \( \int_{a}^{b} (c_1\cdot f(s) + c_2\cdot g(s)) \, ds = c_1\int_{a}^{b} f(s) \, ds + c_2\int_{a}^{b} g(s) \, ds \)

Constants \(c_1\) and \(c_2\) can be factored out of their respective integrals, owing to the fact that they do not vary with \( s \). This gave us:

• \( 2\int_{1}^{2} f(s) \, ds \)
• \( 5\int_{1}^{2} g(s) \, ds \)

Thus, the original complex integral was split into two simpler integrals: one for \(f(s)\) and one for \(g(s)\), each multiplied by its respective constant. This step is crucial for managing intricate function combinations within an integral.

Interval Additivity

Interval additivity is another important property of definite integrals that allows us to add or subtract intervals when integrating a function over a larger domain. This property states that for any function, the integral over a combined interval is the sum of integrals over its sub-intervals.

In our problem, we used interval additivity to calculate \(\int_{1}^{2} g(s) \, ds\). Given:

• \(\int_{0}^{2} g(s) \, ds = 4\)
• \(\int_{0}^{1} g(s) \, ds = -1\)

We know from the property of interval additivity that:

• \(\int_{0}^{2} g(s) \, ds = \int_{0}^{1} g(s) \, ds + \int_{1}^{2} g(s) \, ds\)

This allows us to solve for the unknown integral:

• \(4 = -1 + \int_{1}^{2} g(s) \, ds\)
• \(\Rightarrow \int_{1}^{2} g(s) \, ds = 5\)

Using interval additivity, we could efficiently determine more specific integrals based on known values over different stretches.

Properties of Integrals

Integrals have many useful properties that aid in simplifying mathematical problems, which extend our understanding and applications. In solving the presented exercise, we relied on the following key properties:

• Change of Limits: This property tells us that flipping the limits of integration changes the sign of the integral. For example, \(\int_{2}^{1} f(x) \, dx = -\int_{1}^{2} f(x) \, dx\).
• Zero Integral Property:If a function is integrated over an interval of zero length, the result is zero: \(\int_{a}^{a} f(x) \, dx = 0\).

These properties play a pivotal role in calculations. They offer flexibility by transforming complex expressions and allowing multiple pathways to arrive at a solution, making them an invaluable toolkit in calculus challenges. By employing properties of integrals like these, intricate exercises become more approachable and solvable.