Question

Find the area of the region bounded by the graphs of the equations.$$ y=\sin x, \quad y=\sin ^{3} x, \quad x=0, \quad x=\pi / 2 $$

Short answer

The area of the region bounded by the given curves is \(2 - \frac{\pi}{4}\).

Step-by-step solution

Identify functions and interval

The two curves bounding the area are given by the functions \(y = \sin x\) and \(y = (\sin x)^3\). The interval is given by \(x = 0\) and \(x = \frac{\pi}{2}\).

Set up the integral

The area bounded by the curves can be computed by the definite integral \(\int_{0}^{\frac{\pi}{2}} |\sin x - (\sin x)^3| dx\). However, since \(\sin x \geq (\sin x)^3\) for \(x\) in \([0, \frac{\pi}{2}]\), it is not needed to take the absolute value, and the integral simplifies to \(\int_{a}^{b} (\sin x - (\sin x)^3) dx\).

Evaluate the integral

With the antiderivative of \(\sin x\) being \(-\cos x\) and the antiderivative of \((\sin x)^3\) being \(\frac{1}{3}(\frac{3}{4}x-\frac{1}{2}\sin 2x+\frac{1}{8}\sin 4x)\), the solution of the integral \(\int_{a}^{b} (\sin x - (\sin x)^3) dx\) from \(0\) to \(\frac{\pi}{2}\) can be found by subtracting the antiderivative at the lower limit from the one at the upper, that gives \(-\cos\left(\frac{\pi}{2}\right) + \frac{1}{3}(\frac{3}{4}\cdot\frac{\pi}{2}-\frac{1}{2}\sin(\pi)+\frac{1}{8}\sin(2\cdot\frac{\pi}{2})) - (-\cos(0) + \frac{1}{3}(\frac{3}{4}\cdot 0-\frac{1}{2}\sin(0)+\frac{1}{8}\sin(0)))\).

Simplify the result

The result simplifies to \(2 - \frac{\pi}{4}\).

Key concepts

Integral Calculus

Integral calculus is a fundamental aspect of mathematics that deals with the accumulation of quantities and the area under curves. When faced with a graph, you can use integrals to calculate total values, such as area or volume. Integral calculus is divided into two types: indefinite integrals (antiderivatives) and definite integrals. A definite integral has specific limits and gives a numeric value, which is the case when calculating the area between curves. The process involves finding the antiderivative of a function, and then applying the limits to calculate the final value.

When expressing a definite integral, we write it as \[ \int_{a}^{b} f(x) \, dx \], where \(a\) and \(b\) are the lower and upper limits of the integration, respectively. The function \(f(x)\) is the one we’re integrating over the interval \[a, b\]. The solution to a definite integral represents the net area between the curve and the x-axis over the interval from \(a\) to \(b\).

Area Between Curves

Calculating the area between curves involves finding the region enclosed by the graphs of two functions. The technique is to calculate the integral of the difference between the two functions over a given interval. This is especially useful in comparing the properties or outputs of two different functions that intersect at certain points.

To find the area between the curves of functions \(f(x)\) and \(g(x)\), we use the formula \[ \int_{a}^{b} |f(x) - g(x)| \, dx \], where \(a\) and \(b\) are the points of intersection (or the given interval boundaries). In the case where one function is always on top of the other within the interval, the absolute value can be omitted, and the integral simplifies to \[ \int_{a}^{b} (f(x) - g(x)) \, dx \].

Exercise Improvement Advice

Including visual aids such as graphs to depict the functions and shaded areas representing the region being integrated can significantly aid students in understanding how the integral represents the area between curves. Additionally, discussing the importance of knowing which function is on top within the given interval can clarify why we sometimes do not need the absolute value in our calculations.

Trigonometric Integration

Integrating trigonometric functions is a key technique within calculus. Trigonometric integration involves finding the antiderivatives of trigonometric functions. These integrals can often be solved by using specific trigonometric identities or by employing methods such as substitution.

Some common integrals in trigonometric integration are for functions like \(\sin x\), \(\cos x\), and \(\tan x\), among others. For example, the antiderivative of \(\sin x\) is \( -\cos x\) and that's a recurring element in calculating areas in trigonometric contexts. When dealing with powers of trigonometric functions, such as \(\sin^3 x\), trickier integrals emerge, and decomposition using trigonometric identities might be needed.

Solving Complex Trigonometric Integrals

For more complex integrals, we might also involve formulas like the half-angle identities or the power-reducing formulas to simplify the expressions before integrating. Understanding these techniques is essential for solving integrals involving products or higher powers of trigonometric functions.

Antiderivatives

Antiderivatives are the inverse process of differentiation. If we have a derivative of a function \(f'(x)\), then its antiderivative is the original function \(f(x)\) up to an arbitrary constant \(C\), because derivatives of constants are zero. To denote an antiderivative of a function, we use the integral symbol without limits. For example, an antiderivative of \(f'(x)\) is written as \[ \int f'(x) \, dx = f(x) + C \].

The process of finding antiderivatives is crucial for calculating definite integrals because it involves evaluating the antiderivative at the upper and lower bounds of the integral, which yields the net change of the function over the interval. This concept is formalized by the Fundamental Theorem of Calculus, which connects differentiation and integration.

Role of Antiderivatives in Area Calculation

In the context of area calculations, the antiderivative of a function describes the accumulated area under the curve, starting from a reference point. The area between curves and x-axis over an interval is found by evaluating the antiderivative at both ends of the interval and subtracting these values, as demonstrated in our exercise example. Mastery of antiderivatives is thus imperative for students to effectively tackle definite integral problems in calculus.