Problem 74 - Calculus Single Variable

Question

Find the area of the region bounded by \(y=\sin (x)\) and \(y=\sin ^{2}(x)\) for \(0 \leq x \leq \pi / 2\).

Step-by-Step Solution

Answer

The area bounded by the curves is \( \frac{1}{6} \).

Step 1: Find Points of Intersection

First, we need to find where the functions intersect. Set the equations equal: \( \sin(x) = \sin^2(x) \). It simplifies to \( \sin(x)(1 - \sin(x)) = 0 \). Thus, \( \sin(x) = 0 \) or \( \sin(x) = 1 \). For \( 0 \leq x \leq \pi/2 \), \( x = 0 \) and \( x = \pi/2 \) are the points of intersection.

Step 2: Determine the Functions' Order

Next, we need to establish which function is above the other on the interval. For \( y = \sin(x) \) and \( y = \sin^2(x) \), compare at a point within the interval, say \( x = \pi/4 \). \( \sin(\pi/4) = \sqrt{2}/2 \) and \( \sin^2(\pi/4) = 1/2 \). Thus, \( y = \sin(x) \) is above \( y = \sin^2(x) \) in this interval.

Step 3: Set Up the Integral

The area between the curves can be found using the integral: \[ A = \int_{0}^{\pi/2} (\sin(x) - \sin^2(x)) \, dx \].

Step 4: Simplify the Integral Expression

The integral simplifies to: \[ A = \int_{0}^{\pi/2} \left( \sin(x) - \sin^2(x) \right) \, dx = \int_{0}^{\pi/2} \sin(x) \left( 1 - \sin(x) \right) \, dx \].

Step 5: Use Substitution to Solve Integral

Let \( u = \sin(x) \), therefore \( \frac{du}{dx} = \cos(x) \) and \( du = \cos(x) \, dx \). When \( x = 0 \), \( u = 0 \). When \( x = \pi/2 \), \( u = 1 \). This changes the bounds, and the integral becomes: \[ \int_{0}^{1} u - u^2 \, du \].

Step 6: Evaluate the Integral

Now evaluate the integral: \[ \int_{0}^{1} (u - u^2) \, du = \left[ \frac{u^2}{2} - \frac{u^3}{3} \right]_{0}^{1} \]. Calculating the expression gives \[ \left( \frac{1^2}{2} - \frac{1^3}{3} \right) - \left( \frac{0^2}{2} - \frac{0^3}{3} \right) = \frac{1}{2} - \frac{1}{3} = \frac{1}{6} \].

Step 7: Final Calculation and Conclusion

The integral evaluates to \( \frac{1}{6} \), thus the area between the curves from \( 0 \) to \( \pi/2 \) is \( \frac{1}{6} \) square units.

Key Concepts

Integration by Substitution

Integration by substitution is a crucial technique for solving integrals, especially when you need to simplify an integral involving composite functions. It's very similar to the concept of the chain rule in differentiation. Here, you choose a new variable, usually denoted as \( u \), to substitute a part of the integrand in a way that makes the integration easier.

In the exercise, the integral \( \int (\sin(x) - \sin^2(x))\, dx \) requires simplification. By letting \( u = \sin(x) \), we make the integrand simpler. The differential \( du = \cos(x)\, dx \) helps us replace \( \cos(x)\, dx \) in the original integral.

After substitution, the integral becomes \( \int (u - u^2)\, du \) with altered boundaries according to the function \( u \). This integral is much simpler to evaluate and leads to the straightforward calculation of the area between the curves.

Trigonometric Functions

Trigonometric functions like sine and cosine are fundamental in many mathematical applications, especially in calculus. In the given exercise, you are dealing with \( y = \sin(x) \) and \( y = \sin^2(x) \). These functions periodically oscillate between specific values and have distinct shapes.

The sine function, \( y = \sin(x) \), delivers a smooth wave that runs from -1 to 1. Meanwhile, \( y = \sin^2(x) \) is the square of the sine function, meaning it oscillates between 0 and 1, because the square of any real number will always be non-negative.

While solving such problems, understanding the graphical behavior of these functions helps to determine which function lies above the other in a required interval. In this range, \( [0, \frac{\pi}{2}] \), evaluating both functions at a midpoint like \( \frac{\pi}{4} \) helps confirm their position, guiding the integration process correctly.

Definite Integral

Definite integrals play a crucial role in calculating the exact area under a curve or between two curves over a specific interval. Here, the definite integral is utilized to find the area between two curvy lines, \( y=\sin(x) \) and \( y=\sin^2(x) \), from \( x=0 \) to \( x=\frac{\pi}{2} \).

The definite integral \[\int_{0}^{\frac{\pi}{2}} (\sin(x) - \sin^2(x)) \, dx\]allows you to sum all the tiny stripes between these curves over the given range. This calculation, in essence, is adding the infinite number of differential segments (tiny areas) between these curves from 0 to \( \frac{\pi}{2} \) radians.

Once you apply integration by substitution, the boundaries shift according to your substitution. When you evaluate the definite integral for \( u \), it directly calculates the area, important for physics and engineering problems involving trigonometric functions.