Question

Find the area of the region bounded by the graphs of all four equations: \(f(x)=\sin \left(\frac{x}{2}\right) ; x\) -axis; and the lines, \(x=\frac{\pi}{2}\) and \(x=\pi .\)

Short answer

Answer: The area of the region bounded by the graphs of the four equations is \(\sqrt{2}\).

Step-by-step solution

Identify the intersection points of the equations

We have the following four equations: 1. \(f(x) = \sin\left(\frac{x}{2}\right)\) 2. \(x\)-axis (which is \(y=0\)) 3. \(x = \frac{\pi}{2}\) 4. \(x = \pi\) Since we need to find the area bounded by these equations, we need to determine the intersection points. For the intersection of the function with the \(x\)-axis, set \(f(x)=0\): \(\sin\left(\frac{x}{2}\right) = 0\) Solving this equation leads to \(x = 0, x = 2\pi\). Now we need to check the intersection points with the lines \(x=\frac{\pi}{2}\) and \(x=\pi\): 1. At \(x = \frac{\pi}{2}\), \(f\left(\frac{\pi}{2}\right)=\sin\left(\frac{\pi}{4}\right)\) 2. At \(x = \pi\), \(f(\pi)=\sin\left(\frac{\pi}{2}\right)\) The area we need to find is from \(x=\frac{\pi}{2}\) to \(x=\pi\), since at those points the function is above the \(x\)-axis.

Set up the integral for the area

To find the area bounded by the function and the \(x\)-axis, we set up the integral: \(\text{Area} = \int_{\frac{\pi}{2}}^{\pi} \left|\sin\left(\frac{x}{2}\right)\right| dx\) Since the function is positive in that interval, we don't need to consider the absolute value.

Evaluate the integral

Integrate the expression: \(\text{Area} = \int_{\frac{\pi}{2}}^{\pi} \sin\left(\frac{x}{2}\right) dx\) To solve this integral, we use substitution method. Let \(u=\frac{x}{2}\). Then, \(du=\frac{1}{2} dx\). So the integral becomes: \(\text{Area} = 2 \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \sin u \ du\) Now, integrating with respect to \(u\): \(\text{Area} = 2\left[-\cos u\right]_{\frac{\pi}{4}}^{\frac{\pi}{2}}\) Apply the limits: \(\text{Area} = 2\left[-\cos\left(\frac{\pi}{2}\right)+\cos\left(\frac{\pi}{4}\right)\right]\) \(\text{Area} = 2\left[0 + \frac{\sqrt{2}}{2}\right]\) \(\text{Area} = \sqrt{2}\) So, the area of the region bounded by the graphs of all four equations is \(\sqrt{2}\).

Key concepts

Area under a curve

The concept of finding the area under a curve is a fundamental aspect of calculus. It involves using integration to determine the space between a given function and the x-axis over a specific interval. This technique helps us measure the total accumulation of quantities, such as distance, population, or, in our case, the area.In this exercise, we focus on the area bounded by the function \(f(x) = \sin\left(\frac{x}{2}\right)\) and other specified lines: the x-axis, \(x = \frac{\pi}{2}\), and \(x = \pi\). To find this area, we set up an integral that spans from \(x = \frac{\pi}{2}\) to \(x = \pi\). The correct setup of boundaries ensures we are examining the exact region bounded by these specified lines. This integral will add up the infinitesimal slices of area, giving us the complete area under the curve and above the x-axis within the defined interval.

Trigonometric Integration

Trigonometric integration refers to using trigonometric functions as integrands—functions to be integrated. These types of integrals often appear in problems involving periodic phenomena like waves and oscillations, common in physics and engineering.In this specific problem, our function to integrate is \(\sin\left(\frac{x}{2}\right)\). Integrating sine functions involves techniques often relying on known antiderivatives. In practice, the integral of \(\sin(u)\) with respect to \(u\) is \(-\cos(u)\).When dealing with trigonometric integrals such as these, it is helpful to recognize patterns and connections to their derivatives. Understanding these can greatly simplify the calculation process, as we integrate functions that capture wiggles and curves inherent in periodic functions like sine and cosine.

Substitution Method

The substitution method is a powerful tool in calculus for simplifying the process of integration. It involves changing the variable of integration to make the integrand easier to work with. This technique is particularly useful in cases involving composite functions.For our exercise, integration of \(\sin\left(\frac{x}{2}\right)\) required substitution. We set \(u = \frac{x}{2}\), which simplifies the original integral to be in terms of \(u\), making the integration process straightforward. Furthermore, this change of variable adjusts limits to match the new integral's bounds, now evaluated from \(u = \frac{\pi}{4}\) to \(u = \frac{\pi}{2}\).This method transforms the integral into one that is easier to evaluate—specifically, \(2 \int \sin(u) \, du\). Such manipulations are essential for solving complex integrals efficiently, highlighting the elegance and simplicity substitution offers when tackling such problems.