Question

Find the area of the region bounded by the curves \(y=\sin x\) and \(y=\sin ^{-1} x\) on the interval \(\left[0, \frac{1}{2}\right]\).

Short answer

Answer: The area of the region is \(\frac{\pi}{12}\).

Step-by-step solution

Graph the functions and find intersection points

First, let's graph the two functions \(y=\sin x\) and \(y=\sin^{-1}x\) on the interval \(\left[0, \frac{1}{2}\right]\) and find the points where the two curves intersect. To find the intersection points, we need to solve the equation: $$\sin x = \sin^{-1}x$$ Since \(\sin(0) = 0\) and \(\sin^{-1}(0) = 0\), we have one intersection point at \((0, 0)\). Now, let's find the intersection point at \(x = \frac{1}{2}\). We have \(\sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6}\) and \(\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}\). So the other intersection point is \(\left(\frac{1}{2}, \frac{1}{2}\right)\).

Set up the definite integral

To find the area between the curves, we need to integrate the difference of their \(y\) values: $$\int_{a}^{b} (\sin^{-1}x - \sin x) dx$$ We have found the intersection points \((0,0)\) and \(\left(\frac{1}{2}, \frac{1}{2}\right)\). We'll use these as the integration limits, so we have \(a=0\) and \(b=\frac{1}{2}\). Now we have: $$\int_{0}^{\frac{1}{2}} (\sin^{-1}x - \sin x) dx$$

Evaluate the definite integral

To evaluate the integral, we will use integration by parts, with \(u=\sin^{-1}x\) and \(dv = -\sin{x}dx\): $$\int (\sin^{-1}x) (-\sin x) dx = -\int u dv$$ Now, we differentiate \(u\) and integrate \(dv\): $$du = \frac{1}{\sqrt{1-x^2}} dx, \ v = \int{-\sin x dx} = \cos{x}$$ Now the integration by parts formula gives us: $$-\int u dv = -uv + \int v du$$ Plugging in values, we get: $$-\int_{0}^{\frac{1}{2}} (\sin^{-1}x) (-\sin x) dx = \left[-(\sin^{-1}x)\cos x + \int_{0}^{\frac{1}{2}}\frac{\cos x}{\sqrt{1-x^2}} dx\right]$$ We can evaluate this integral by making a substitution. Let \(x=\sin{t}\), so \(dx = \cos{t} dt\). Our integral becomes: $$\int_{0}^{\frac{\pi}{6}}\frac{\cos{t}}{\sqrt{1-\sin^2{t}}} \cos{t} dt = \int_{0}^{\frac{\pi}{6}}\cos^2{t} dt$$ Now, we can use the identity \(\cos^2{t} = \frac{1 + \cos(2t)}{2}\) to evaluate the integral: $$\int_{0}^{\frac{\pi}{6}}\cos^2{t} dt = \int_{0}^{\frac{\pi}{6}}\frac{1 + \cos(2t)}{2} dt = \frac{1}{2}\int_{0}^{\frac{\pi}{6}}(1 + \cos(2t))dt$$ Separating the terms, we have: $$= \frac{1}{2}\left[\int_{0}^{\frac{\pi}{6}}dt + \int_{0}^{\frac{\pi}{6}}\cos(2t)dt\right]$$ Evaluating each integral: $$= \frac{1}{2}\left[t \Big|_0^{\frac{\pi}{6}} + \frac{1}{2}\sin(2t)\Big|_0^{\frac{\pi}{6}}\right] = \frac{1}{2}\left[\frac{\pi}{6} + 0\right] = \frac{\pi}{12}$$ Putting the integration by parts result back together: $$-\left[-(\sin^{-1}x)\cos x + \int_{0}^{\frac{1}{2}}\frac{\cos x}{\sqrt{1-x^2}} dx\right]_{0}^{\frac{1}{2}} = -\left[-\frac{\pi}{12} - 0\right] = \frac{\pi}{12}$$

Final answer

The area of the region bounded by the curves \(y=\sin x\) and \(y=\sin^{-1}x\) on the interval \(\left[0, \frac{1}{2}\right]\) is: $$A = \frac{\pi}{12}$$

Key concepts

Area between curves

The area between two curves is a crucial concept in calculus that involves finding the space between the graphs of two functions. This is often done over a specified interval. To find this area, we subtract the function values of the lower curve from the upper curve.
For instance, if you have two functions, such as the sine function, \(y = \sin{x}\), and its inverse, \(y = \sin^{-1}x\), the area would be determined by integrating the difference on the given interval.
The formula to find the area between the curves \(y = f(x)\) and \(y = g(x)\) from \(x = a\) to \(x = b\) is given by:

• \( \int_{a}^{b} (f(x) - g(x)) dx \)

Here, \( f(x)\) is the function representing the upper curve, and \(g(x)\) is representing the lower curve on the interval \([a, b]\). It's important to identify which function is on top and which is at the bottom in the specified range for accurate calculations.

Integration by parts

Integration by parts is a technique used to solve integrals that are products of two functions, making it ideal for functions like the inverse trigonometric and trigonometric products in this exercise. This method can be derived from the product rule for differentiation and provides an excellent strategy to tackle these complex integrals.
The integration by parts formula is given by:

• \( \int u \, dv = uv - \int v \, du \)

Here, \(u\) is a function we will differentiate, and \(dv\) is a function we will integrate. It is crucial to choose these elements wisely to simplify the problem.
In our solution to find the area bounded by the curves, we selected \( u = \sin^{-1}x \) and \( dv = -\sin{x} dx \). Then, we differentiated \( u \) and integrated \( dv \) as part of this method, making it easier to solve the integral appropriately.

Definite integrals

A definite integral is a type of integral that computes the accumulation of quantities between two specified limits. In this exercise, we want to find the area between two curves over a given interval, making definite integrals the ideal tool.
The limits of integration indicate the start and end points on the \(x\)-axis. For our exercise, these are \(0\) and \(\frac{1}{2}\).
The general formula for a definite integral from \(a\) to \(b\) is:

• \( \int_{a}^{b} f(x) \, dx \)

This notation signifies the area under \( f(x)\) from \(a\) to \(b\). In our case, we want \(f(x) = \sin^{-1}x - \sin x\) to find the accumulated area between the two given curves.

Inverse trigonometric functions

Inverse trigonometric functions form an integral part of calculus, providing solutions to equations involving trigonometric functions. For example, the inverse sine function, denoted as \( \sin^{-1}x \), returns the angle whose sine is \(x\). This is useful in the exercise as we need to calculate where the inverse function interacts with the trigonometric functions to find intersections.
In calculus, inverse trigonometric functions have derivatives that are particularly helpful in integration. For instance, the derivative of \( \sin^{-1}x \) is given by

• \( \frac{1}{\sqrt{1-x^2}} \)

This particular derivative was used in integration by parts to find the area bounded by the curves \(y=\sin x\) and \(y=\sin^{-1}x\). Knowing the properties of inverse trigonometric functions can greatly aid in solving complex calculus problems seamlessly.