Question

Find the area of the region bounded by the graphs of the equations \(y=\sin x, y=\sin ^{3} x, x=0\), and \(x=\frac{\pi}{2}\).

Short answer

The area of the region is \(\frac{3}{4}\).

Step-by-step solution

Understanding the Problem

We need to find the area of the region bounded by the graphs of two functions: \(y = \sin x\) and \(y = \sin^3 x\) between the vertical lines \(x = 0\) and \(x = \frac{\pi}{2}\). This requires using calculus to conduct integration.

Identifying the Limits of Integration

The region of interest lies between \(x = 0\) and \(x = \frac{\pi}{2}\). Therefore, these will be the limits for our definite integral calculations.

Determining Subtraction Function

Since \(\sin x\) is greater than \(\sin^3 x\) over the interval \(0 \leq x \leq \frac{\pi}{2}\), we need to integrate \(\sin x - \sin^3 x\). This expresses the vertical difference between the two curves.

Evaluating the Integral

Now, compute the integral \[ \int_0^{\frac{\pi}{2}} (\sin x - \sin^3 x) \, dx.\] Breaking it down, the integral becomes \[ \int_0^{\frac{\pi}{2}} \sin x \, dx - \int_0^{\frac{\pi}{2}} \sin^3 x \, dx.\]

Calculate Integral of \(\sin x\)

The integral of \(\sin x\) can be solved as \(-\cos x\). Evaluate this from \(0\) to \(\frac{\pi}{2}\):\[ \int_0^{\frac{\pi}{2}} \sin x \, dx = [-\cos x]_0^{\frac{\pi}{2}} = -\cos\left(\frac{\pi}{2}\right) - (-\cos(0)) = 1.\]

Calculate Integral of \(\sin^3 x\) using Substitution Method

Use the substitution \( u = \sin x\), then \( du = \cos x \, dx\). Thus, the integral \[ \int \sin^3 x \, dx\] becomes\[ \int u^3 du = \frac{u^4}{4}.\]Change back to \((\sin x)^4\) and evaluate from \(0\) to \(\frac{\pi}{2}\):\[ \left[ \frac{(\sin x)^4}{4} \right]_0^{\frac{\pi}{2}} = \frac{1^4}{4} - \frac{0^4}{4} = \frac{1}{4}.\]

Calculating the Area

Finally, subtract the result of the integrals via the formula:\[ \text{Area} = 1 - \frac{1}{4} = \frac{3}{4}.\]

Verification

Check calculations for each part and ensure that the subtraction and integral evaluation are correct.

Key concepts

Definite Integral

A definite integral is a fundamental concept in calculus that represents the accumulation of a quantity, like area, between a curve and the x-axis over a specific interval. In simple terms, it is used to calculate the total amount of something, like area or volume, within a given set of boundaries. For the problem at hand, the definite integral helps us find the area between two curves by providing a structured method to add up tiny slices of area over the interval from 0 to \(\frac{\pi}{2}\).

To solve the exercise, we set up a definite integral that measures the area between the curves \(y = \sin x\) and \(y = \sin^3 x\). These boundaries are defined by the vertical lines \(x = 0\) and \(x = \frac{\pi}{2}\), so our integral becomes:

• \[ \int_0^{\frac{\pi}{2}} (\sin x - \sin^3 x) \, dx \]

This integral takes into account the slicing approach, breaking down the area into small segments and summing them up within the specified limits.

Area Between Curves

When asked to find the area between curves, what we're really doing is finding the difference between the values of two functions over a certain range. Think of it like peeling an orange. You have the outer layer and the inner juicy part. You want to measure the space between those two layers. But in math, the curves can represent more complex relationships.

In our exercise, we examine the area between \(y = \sin x\) and \(y = \sin^3 x\). The integral expression \(\int_0^{\frac{\pi}{2}} (\sin x - \sin^3 x) \, dx\) handles this by computing the difference in height between two curves throughout the interval from \(x = 0\) to \(x = \frac{\pi}{2}\).

• Always subtract the lower curve from the higher one over the interval.
• Calculate each integral separately when the expression comprises more than one term.

In this problem, we find the integral of \(\sin x\) and subtract the integral of \(\sin^3 x\) to determine the total area enclosed by those curves.

Trigonometric Functions

Trigonometric functions like \(\sin x\) and \(\sin^3 x\) are essential for modeling periodic phenomena like sound waves, light, and oscillations. These are functions of an angle and are used commonly in calculus for integration and area calculations between curves.

For our problem, understanding \(\sin x\) and its powers is crucial. The function \(y = \sin x\) represents a wave-like pattern starting at the origin and moving upwards and downwards until it reaches \(\frac{\pi}{2}\). Meanwhile, \(y = \sin^3 x\) is a modification that dampens the wave more considerably, hugging closer to the x-axis but mimicking the general wave-like pattern of the \(\sin\) function.

• The amplitude of \(\sin x\) determines its peak height at \(1\).
• \(\sin^3 x\) maintains the same roots, but with a reduced peak.

Grasping these trigonometric insights helps us appreciate why \(\sin x\) surpasses \(\sin^3 x\) in height except at specific junctures, thus forming the basis to calculate the area between them.