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Chapter 6: Problem 75

A region \(\mathcal{R}\) is described. Calculate its area. $$ \begin{aligned} &\mathcal{R} \text { is the region that is bounded above by } y=\pi x / 2 \text { and }\\\ &\text { below by } y=\arcsin (x) \text { for } 0 \leq x \leq 1 \end{aligned} $$

Short Answer

Expert verified
The area of region \( \mathcal{R} \) is \( 1 - \frac{\pi}{4} \).

Step by step solution

01

Identify the Functions

The region \( \mathcal{R} \) is bounded above by the line \( y = \frac{\pi x}{2} \) and below by the curve \( y = \arcsin(x) \). We will compute the area between these two curves.
02

Set Up the Integral for Area

The area \( A \) of region \( \mathcal{R} \) between the two curves from \( x = 0 \) to \( x = 1 \) is given by the integral \[ A = \int_{0}^{1} \left( \frac{\pi x}{2} - \arcsin(x) \right) \, dx \].
03

Integrate the Function

Calculate the integral \[ \int_{0}^{1} \frac{\pi x}{2} \, dx \] which equals \( \frac{\pi}{2} \left[ \frac{x^2}{2} \right]_{0}^{1} = \frac{\pi}{4} \).
04

Integrate the Second Function

Calculate the integral of \( \arcsin(x) \) with respect to \( x \), which equals \[ \int_{0}^{1} \arcsin(x) \, dx = x \arcsin(x) + \sqrt{1-x^2} + C \bigg|_{0}^{1} = \frac{\pi}{2} - 1 \].
05

Compute the Area

The area \( A \) is then \( \frac{\pi}{4} - (\frac{\pi}{2} - 1) = 1 - \frac{\pi}{4} \).
06

Simplify the Result

Thus, the area of the region \( \mathcal{R} \) is given by \( 1 - \frac{\pi}{4} \) which approximates to approximately \( 0.215 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Definite Integrals
Definite integrals are a fundamental concept in calculus that allow us to calculate the accumulation of quantities, like finding the area under a curve. The primary goal of using a definite integral is to evaluate a function over a specific interval. In simple terms, it helps us to measure the space between the curve (or line) and the horizontal x-axis within the interval we are considering.

Let's delve deeper into this:
  • The limits of integration tell us the interval over which we are finding the area. In this case, the interval is from 0 to 1.
  • The integral symbol \( \int \) represents the process of integration鈥攁dding up infinite small pieces to find the total.
  • The result gives a number representing the accumulated change, in this scenario, the area between the curves.
By understanding definite integrals, you have the mathematical toolset needed to compute areas between curves or understand accumulated quantities in various fields.
Calculating Area Between Curves
The area between curves is a classic applications of definite integrals. In this context, it helps us determine the region between two given functions over a certain interval. The exercise you encountered requires calculating the area of a region \( \mathcal{R} \) that is bounded by the curves \( y = \frac{\pi x}{2} \) on top and \( y = \arcsin(x) \) at the bottom.

Here鈥檚 how this works:
  • First, recognize which function is on top (here, \( y = \frac{\pi x}{2} \)) and which is below (here, \( y = \arcsin(x) \)).
  • The area \( A \) is computed by subtracting the lower function from the upper function, then integrating over the given interval.
  • The integral expression \( \int_{0}^{1} \left( \frac{\pi x}{2} - \arcsin(x) \right) \, dx \) is set up to calculate this difference in height across the entire range from 0 to 1.
By effectively computing these activities step-by-step, the calculation of the area becomes more structured and logical.
Exploring Trigonometric Functions
Trigonometric functions, like \( \arcsin \), play a crucial role in various mathematical computations, especially involving curves and angles. In our exercise, the functional form \( y = \arcsin(x) \) describes part of the boundary of region \( \mathcal{R} \). Understanding how these functions behave is vital for accurate problem solving.

Some key insights about \( \arcsin(x) \):
  • \( \arcsin(x) \) is the inverse function of \( \sin(x) \), which means it tells us the angle whose sine is \( x \).
  • This function is defined for \( x \) values between -1 and 1, ensuring the result is a real number.
  • In the context of calculus, \( \arcsin(x) \) is significant due to its non-linear shape and its integration properties.
Knowing these fundamentals helps in understanding the curve's trajectory, allowing us to visualize and compute areas more accurately and effectively. With this, you can see how trigonometric functions contribute to calculating real-world shapes and areas.

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Most popular questions from this chapter

By making the substitution \(u=\sin (x),\) show that $$\int \sin (x) \cos (x) d x=\frac{1}{2} \sin ^{2}(x)+C$$ By making the substitution \(u=\cos (x),\) show that $$\int \sin (x) \cos (x) d x=-\frac{1}{2} \cos ^{2}(x)+C $$By using the formula for \(\sin (2 x)\), show that $$\int \sin (x) \cos (x) d x=-\frac{1}{4} \cos (2 x)+C$$ Explain why these answers are not contradictory.

Let \(b\) be the abscissa of the first-quadrant point of intersection of the graphs of \(y=x^{2}\) and \(y=\frac{3 x^{3}+4 x}{x^{4}+3 x^{2}+2}\) Calculate the area of the region between the two curves for \(0 \leq x \leq b\).

Evaluate the given definite integral. \(\int_{1}^{2} \frac{7 x^{2}+6}{x^{2}\left(x^{2}+1\right)} d x\)

Evaluate the given definite integral. \(\int_{1}^{2} \frac{2 x^{2}-x+2}{x\left(x^{2}+1\right)} d x\)

In each of Exercises \(63-68\), use the Comparison Theorem to establish that the given improper integral is convergent. $$ \int_{1}^{\infty} \frac{x}{1+x^{3}} d x $$
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