Problem 25 - Calculus Single Variable

Question

Compute the average value of \(f\) over \([a, b]\). \(f(x)=\cos (x) \quad a=0, b=\pi / 2\)

Step-by-Step Solution

Answer

The average value of \(f(x) = \cos(x)\) over \([0, \frac{\pi}{2}]\) is \(\frac{2}{\pi}\).

Step 1: Understand the Formula for Average Value of a Function

The average value of a continuous function \(f\) over an interval \([a, b]\) is given by the formula: \[ f_{avg} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx \] For this problem, \(f(x) = \cos(x)\), \(a = 0\), and \(b = \frac{\pi}{2}\).

Step 2: Apply the Formula with Given Values

Substitute \(f(x) = \cos(x)\), \(a = 0\), and \(b = \frac{\pi}{2}\) into the formula: \[ f_{avg} = \frac{1}{\frac{\pi}{2} - 0} \int_{0}^{\frac{\pi}{2}} \cos(x) \, dx \] This simplifies to: \[ f_{avg} = \frac{2}{\pi} \int_{0}^{\frac{\pi}{2}} \cos(x) \, dx \]

Step 3: Evaluate the Integral

Calculate the definite integral: \[ \int_{0}^{\frac{\pi}{2}} \cos(x) \, dx \] The antiderivative of \(\cos(x)\) is \(\sin(x)\), so evaluate: \[ \sin(x) \Big|_{0}^{\frac{\pi}{2}} = \sin\left(\frac{\pi}{2}\right) - \sin(0) \] Simplifies to: \[ 1 - 0 = 1 \]

Step 4: Compute the Average Value

Substitute the value of the integral back into the average value formula: \[ f_{avg} = \frac{2}{\pi} \cdot 1 = \frac{2}{\pi} \] Thus, the average value of \(f(x) = \cos(x)\) over the interval \([0, \frac{\pi}{2}]\) is \(\frac{2}{\pi}\).

Key Concepts

Definite Integral

A definite integral is a fundamental concept in calculus that represents the accumulation of quantities over an interval. It is used to find areas under curves, total accumulated values, and other applications where sum aggregation over a range is required. When we calculate the definite integral of a function, we are essentially finding the net area between the function and the x-axis over the given interval.
It is expressed as:

• The symbol \( \int \) indicates integration.
• The limits of integration \( a \) and \( b \) specify the interval.
• The expression \( f(x) \, dx \) indicates the function being integrated, with \( dx \) representing an infinitesimally small change in \( x \).

For example, in the exercise, the definite integral \( \int_{0}^{\frac{\pi}{2}} \cos(x) \, dx \) computes the total area under the cosine function from 0 to \( \frac{\pi}{2} \). For the average value problem, we use this computation to determine how the values of \( f(x) \) are distributed within the interval.

Trigonometric Functions

Trigonometric functions, like sine and cosine, are essential in calculus for modeling periodic phenomena. Their behavior is predictable due to their periodic nature, meaning they repeat values at regular intervals. In the context of the integration problem, we focus on the cosine function, which has a few key characteristics worth noting:

• The cosine function, \( \cos(x) \), oscillates between -1 and 1.
• It is an even function, meaning \( \cos(-x) = \cos(x) \), which is symmetric about the y-axis.
• Cosine has a period of \(2\pi\), meaning its pattern repeats every \(2\pi\) units.

These characteristics are beneficial when calculating definite integrals because understanding where and how often a function crosses the axis can simplify calculations and offer insight into the behavior of the function over an interval.
In our example, using this knowledge helps us evaluate the integral of \( \cos(x) \) over \( [0, \frac{\pi}{2}] \) efficiently.

Antiderivatives

Finding an antiderivative of a function is crucial when solving problems involving definite integrals. An antiderivative of a function \( f(x) \) is another function \( F(x) \) such that the derivative of \( F(x) \) is \( f(x) \). In essence, we reverse the differentiation process to find the function that gave us \( f(x) \) when differentiated.
For example, in the average value calculation, we needed the antiderivative of \( \cos(x) \), which is \( \sin(x) \). Knowing this allows us to evaluate the definite integral:

• To find \( \int_{a}^{b} \cos(x) \, dx \), we find an antiderivative, \( \sin(x) \).
• We then evaluate this from \( 0 \) to \( \frac{\pi}{2} \) to find \( \sin\left( \frac{\pi}{2} \right) - \sin(0) = 1 - 0 \).

Antiderivatives play a vital role in calculating the average value of a function over an interval, as they help us determine the exact accumulation of value.
This makes it possible to apply the fundamental theorem of calculus, bridging antiderivatives and definite integrals effectively.