Problem 31 - Calculus for Biology and Medicine

Question

Find the areas of the regions bounded by the lines and curves. \(y=\sin x, y=1\) from \(x=0\) to \(x=\frac{\pi}{4}\)

Step-by-Step Solution

Answer

The area is \(\frac{\pi}{4} + \frac{\sqrt{2}}{2} - 1\).

Step 1: Graph the Functions

First, graph the functions of interest, which are the sine function, \(y = \sin x\), and the horizontal line, \(y = 1\), over the given interval, \(x \in [0, \frac{\pi}{4}]\). This will help you visualize the bounded region whose area you need to calculate.

Step 2: Identify the Bounded Region

Determine the area enclosed between the curve \(y = \sin x\) and the line \(y = 1\). This means identifying which curve is above the other in the interval. On \([0, \frac{\pi}{4}]\), \(y = 1\) is above \(y = \sin x\).

Step 3: Write the Integral Expression

Set up the definite integral that represents the area between the two curves. The integral is formed by subtracting the lower curve, \(y = \sin x\), from the upper curve, \(y = 1\). The expression for the area is: \[ A = \int_{0}^{\frac{\pi}{4}} (1 - \sin x) \, dx \]

Step 4: Calculate the Integral

Evaluate the integral to find the area. First, integrate each term separately:* The integral of \(1\) with respect to \(x\) is \(x\).* The integral of \(\sin x\) with respect to \(x\) is \(-\cos x\).Substitute the limits of integration into the antiderivative:\[ A = \left[ x + \cos x \right]_{0}^{\frac{\pi}{4}} = \left( \frac{\pi}{4} + \cos \frac{\pi}{4} \right) - \left( 0 + \cos 0 \right) \]

Step 5: Evaluate and Simplify

Substitute the values of \(\cos \, \frac{\pi}{4} = \frac{\sqrt{2}}{2}\) and \(\cos \, 0 = 1\) into the expression:\[ A = \left( \frac{\pi}{4} + \frac{\sqrt{2}}{2} \right) - (0 + 1) \]\[ A = \frac{\pi}{4} + \frac{\sqrt{2}}{2} - 1 \]

Key Concepts

Integral Calculus

Integral Calculus is a fundamental part of calculus, focusing on the accumulation of quantities and the area under curves. Think of it as a comprehensive tool to calculate spaces under curves on a given interval. Here are some key points about Integral Calculus:

• It revolves around the concept of an integral, which is essentially a continuous sum.
• Integrals can be indefinite, representing a family of functions, or definite, calculating specific values like area.

To understand how we use integral calculus in practice, consider the function given in the exercise: determinations like this involve setting up and evaluating integrals to find the areas between curves and lines over specific intervals. The process involves a precise calculation which informs us of how much space curves occupy between two boundaries.

Definite Integral

The definite integral is a crucial concept when it comes to calculating the exact value of the area under a curve. It is expressed and evaluated over a specific interval, so here's what you need to know:

• A definite integral is depicted as \ \( \int_{a}^{b} f(x) \, dx \ \), where \( a \) and \( b \) are the limits indicating the range of integration.
• The outcome of a definite integral is a numerical value, often representing scenarios like total distance over time or area.

In the context of the original problem, the definite integral was set up as \[ \int_{0}^{\frac{\pi}{4}} (1 - \sin x) \, dx \], which calculates the area between the sine wave and a horizontal line within the top limit at \( x = \frac{\pi}{4} \). Here, evaluating the integral provides a precise measurement of the space between the two functions on the given interval, crucial in geometric problems like the one provided.

Area between Curves

Finding the area between curves is a fascinating application of calculus that allows us to measure the space enclosed by different functions. This application becomes particularly interesting when the curves intersect or form a region. Here is a simplified breakdown of how this is often approached:

• First, visualize both curves on the graph, identifying which function sits above the other over the specified interval.
• Set up an integral expression that represents the difference between the upper function and the lower function.

The process was illustrated in the exercise by graphing the sine curve and the line \( y = 1 \), then determining the area between them on \([0, \frac{\pi}{4}]\). It involved evaluating the integral \[ \int_{0}^{\frac{\pi}{4}} (1 - \sin x) \, dx \]. By executing these steps, you can find the exact area of the region, a critical step for problems dealing with spatial reasoning and various real-world applications.