Problem 5 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Find the area in the first quadrant bounded by \(y=\arcsin x, y=\pi / 2\), and \(x=0 .\) (Hint: To get an exact answer it will be simplest to integrate with respect to \(y .\).)

Step-by-Step Solution

Answer

The area of the bounded region in the first quadrant is 1.

Step 1: Understanding the equations and plot them

In a typical Cartesian plane, plotting these three equations \(y=\arcsin x, y=\pi / 2, x=0\) will reveal the area bounded by them in the first quadrant. This clear visual representation helps to plan the limits of integration and understand the nature of the area in question.

Step 2: Inverse \(\arcsin x\) and setup the integral

Given the function \(y=\arcsin x\), \(x\), can be written as \(\sin y\) by inverting the function. Therefore, integrating with respect to \(y\), the integral will be set up this way: \(\int_{0}^{\pi / 2} \sin y dy\). The limits came from \(y=0\) (corresponding to \(x=0\)) and \(y=\pi / 2\) (given directly in the problem).

Step 3: Evaluate the definite integral

Now evaluate the definite integral. The antiderivative of \(\sin y\) is \(-\cos y\), and remember to apply these to upper and lower limit. So \(-\cos y \Bigg|_{0}^{\pi / 2} = -\cos(\pi / 2) - (-\cos(0)) = 0 - (-1) = 1.\)

Key Concepts

Area under curve

Calculating the area under a curve is a fundamental concept in calculus. When given a function and certain boundaries, we often need to compute the area contained within these limits to find meaningful solutions to real-world problems. This task is achieved using the definite integral, which allows us to sum infinitely small slices under the curve, bounded by the specified limits.

In this particular problem, the area is determined under the curve of the function \(y = \arcsin x\) from \(y = 0\) to \(y = \pi/2\), and bounded on the left by the line \(x = 0\). The function \(y = \arcsin x\) is an inverse trigonometric function, which makes finding the area more intriguing as it involves integration with respect to \(y\).

When visualized on a graph, these equations form a suitable region in the first quadrant where the area calculation needs to occur. Using definite integrals with these boundaries, we precisely determine the space under the curve by highlighting the beginning and ending limits.

First quadrant

The first quadrant in the Cartesian coordinate system is defined as the region where both \(x\) and \(y\) are positive. This is important for defining the area we seek to find. When solving problems relating to the first quadrant, it sets a clear boundary that ensures solutions are restricted to positive quadrants solely.

In the exercise, all the bounds for the calculation, such as \(y = \arcsin x\), \(y = \pi/2\), and \(x = 0\), help define a closed segregated area in this first quadrant. Proper understanding of these confines is crucial, as any area outside of this quadrant does not contribute to the solution.

Being aware of these constraints helps to simplify the setup of the integral in this scenario, and it provides a strong foundation for visualizing problems that involve quadrant-specific areas during integral evaluations.

Inverse trigonometric functions

Inverse trigonometric functions, such as arcsine, play a vital role in both pure and applied mathematics. These functions allow us to find the angle with a specific trigonometric value, essentially reversing the default trigonometric outcomes.

In the exercise, we come across \(y = \arcsin x\). When finding the area under curves involving such functions, it's often practical to invert them where possible. For instance, we write \(x = \sin y\) from \(y = \arcsin x\), to simplify calculations by integrating with respect to \(y\) instead of \(x\). This transforms the problem into a more solvable form as seen in the solution provided.

The inverse function allows us to easily identify and evaluate the limits of integration, making them immensely valuable when dealing with arcsine, arccosine, and other inverse trigonometric functions in calculus-related queries. Understanding and skillfully handling these conversions is key for integrating functions involving inverse trigonometric expressions.