Problem 30 - Calculus for Biology and Medicine

Question

Find the areas of the regions bounded by the lines and curves. \(y=1, y=\cos x\) from \(x=0\) to \(x=\frac{\pi}{2}\)

Step-by-Step Solution

Answer

The area is \( \frac{\pi}{2} - 1 \).

Step 1: Understand the Problem

We need to find the area between the line \(y = 1\) and the curve \(y = \cos x\) from \(x=0\) to \(x=\frac{\pi}{2}\). The area is located between these two graphs for the specified range on the x-axis.

Step 2: Set Up the Integral

The area between two curves \(y = f(x)\) and \(y = g(x)\) is given by the integral \(\int_{a}^{b} [f(x) - g(x)] \, dx\). Here, \(f(x) = 1\) and \(g(x) = \cos x\), with limits of integration from \(x = 0\) to \(x = \frac{\pi}{2}\). Set up the integral: \[ \int_{0}^{\frac{\pi}{2}} [1 - \cos x] \, dx \]

Step 3: Integrate the Function

Evaluate the integral \(\int [1 - \cos x] \, dx\). This results in the two integrals: \[ \int 1 \, dx = x \quad \text{and} \quad \int \cos x \, dx = \sin x \].Thus, the evaluated integral becomes: \[ \left[ x - \sin x \right]_{0}^{\frac{\pi}{2}} \].

Step 4: Evaluate the Integral

Substitute the upper and lower limits into the evaluated integral:\[ \left[ \frac{\pi}{2} - \sin\left(\frac{\pi}{2}\right) \right] - \left[ 0 - \sin(0) \right] \].Since \(\sin(\frac{\pi}{2}) = 1\) and \(\sin(0) = 0\), the expression simplifies to:\[ \frac{\pi}{2} - 1 - 0 \].

Step 5: Simplify the Result

Calculate the final value by simplifying:\( \frac{\pi}{2} - 1 \).

Key Concepts

Area Between Curves

When you're trying to find the area between two curves, visualize the space between them on a graph. This area represents a region bounded by the curves.
For example, given the line and curve in our problem, you're finding the area between the horizontal line at the top, and the wavy curve below it.

• Here, the top boundary is the line: \(y = 1\), a simple horizontal line across the graph.
• The bottom boundary is given by the curve: \(y = \cos x\).

To find the area between these two, you calculate it using a definite integral. The curves intersect at the specified points which define your limits of integration, from \(x = 0\) to \(x = \frac{\pi}{2}\). By integrating the difference between these functions within the limits, you successfully compute the area between them.

Definite Integrals

Definite integrals are a fundamental concept in integral calculus. They help calculate the exact area under a curve between two points.
The integral evaluates the accumulation of quantities, effectively giving an area measure.
For the exercise, you used the formula: \[\int_{a}^{b} [f(x) - g(x)] \, dx\]Where \(f(x)\) is the upper function (in our example \(1\)) and \(g(x)\) is the lower function (here \(\cos x\)).
This setup ensures you find the net area between these curves over the interval from \(x = 0\) to \(x = \frac{\pi}{2}\).

• The integration process transforms the symbolic expression into actual values. First, you find an antiderivative, and then you evaluate this at the upper and lower limits of the interval.
• The difference of these values gives you the total area between the curves.
  In this case, the steps led to a solution of \(\frac{\pi}{2} - 1\), which represents the bounded region's exact area.

Trigonometric Functions

Trigonometric functions, like sine and cosine, are crucial in various integral calculus problems. They describe periodic phenomenons and their graphs have distinct wave-like shapes.
For instance, in our example, \(\cos x\) is the trigonometric function forming the lower boundary of the area.
Understanding how these functions behave is key.

• The function \(\cos x\) oscillates between -1 and 1, with specific values at crucial points, such as \(\cos 0 = 1\) and \(\cos \frac{\pi}{2} = 0\).
• These values are part of what determines the area between the lines.

Knowing certain identities and integrals involving these functions is very helpful.
For example, recognizing basic integrals: \[\int \cos x \, dx = \sin x\]By applying this, you correctly evaluate the integral necessary to find the area between the defined curves.