Question

Areas of regions Find the area of the region bounded by the graph of \(f\) and the \(x\) -axis on the given interval. $$f(x)=\sin x \text { on }\left[-\frac{\pi}{4}, \frac{3 \pi}{4}\right]$$

Short answer

Answer: 2

Step-by-step solution

Split the Integral

Since \(\sin x\) is negative on the interval \(\left[-\frac{\pi}{4},0\right]\) and positive on the interval \(\left[0,\frac{3\pi}{4}\right]\), we will split the integral into two parts. The first part will be the value of \(\sin x\) on the interval \(\left[-\frac{\pi}{4},0\right]\) and then change its sign as adding the area of \(\sin x\) on the interval \(\left[0,\frac{3\pi}{4}\right]\): $$\text{Area} = \left|- \int_{-\frac{\pi}{4}}^{0} \sin x \, dx\right| + \int_{0}^{\frac{3\pi}{4}} \sin x \, dx$$

Integrate \(\sin x\)

Now, integrate \(\sin x\) twice, once for each part of the integral: $$\text{Area} = \left|- \left[ -\cos x\right]_{-\frac{\pi}{4}}^{0}\right| + \left[ -\cos x\right]_{0}^{\frac{3\pi}{4}}$$

Evaluate the integrals

Next, plug in the bounds of the intervals and calculate the area: $$\text{Area} = \left|-(-\cos{0} + \cos{(-\frac{\pi}{4})})\right| + (-\cos{\frac{3\pi}{4}}-\cos{0})$$

Simplify the expression

Finally, simplify the expression and find the area using the values of the cosines: $$\text{Area} =\left|-(-1 + \frac{\sqrt{2}}{2})\right| + \left(-\frac{-\sqrt{2}}{2}-1\right)$$ $$\text{Area} = 1-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}+1$$ $$\text{Area} = 2$$ So, the area of the region bounded by the graph of \(f(x)=\sin x\) and the \(x\)-axis on the given interval is 2.

Key concepts

Definite Integral

A definite integral represents the signed area under a curve between two specified points on the x-axis. In this problem, the integral of the function \( f(x) = \sin x \) over the interval \([-\frac{\pi}{4}, \frac{3\pi}{4}]\) is computed. In practice, this involves calculating the area of the region that lies below the curve and above the x-axis, as well as any regions above the curve and below the x-axis.
The definite integral is denoted as \( \int_a^b f(x) \, dx \), where \( a \) and \( b \) are the limits of integration. It provides a way to aggregate and simplify the addition of infinitely many infinitesimal parts between \( a \) and \( b \).

• The result of a definite integral gives a numerical value representing this total accumulation of area.
• Importantly, because we are calculating a definite integral, the result includes taking into account the direction of the area (whether it is above or below the x-axis).

Understanding how definite integrals work is key to solving problems involving areas under curves, such as the one presented here.

Trigonometric Functions

Trigonometric functions such as \(\sin\), \(\cos\), and \(\tan\) play a crucial role in various mathematical analyses, including calculus. In this exercise, we focus on the sine function, \( f(x) = \sin x \).
The sine function has properties that make it periodic with a known range of values, fluctuating between \(-1\) and \(1\). It's a smooth, continuous wave-like function that is commonly used in these types of integrals because of its cyclical nature.

• When dealing with integrals and sine functions, you often leverage symmetry and periodicity.
• In this example, \(\sin x\) is split across an interval that contains both negative and positive values, which requires treating each segment separately to ensure positive areas are always considered.

The specific choice of intervals from the exercise taps into the positive and negative halves of the sine wave, leading to the calculated areas that are a result of balanced integration.

Area under Curve

Finding the area under a curve is often the primary goal when solving integrals in calculus. The area under the curve of a function like \( f(x) = \sin x \) between two points helps determine the actual physical region bounded by the graph and the x-axis.
In order to solve for the area effectively, we need to consider:

• Which segments of the function are above the x-axis and contribute positively to the area.
• Which segments are below the x-axis and must be considered with care to ensure contributions are added as positive values. This often involves flipping the sign.

In this problem, the curve crosses the x-axis which necessitates splitting the integral to correctly account for all areas. By integrating across the chosen intervals and ensuring any lower segment's contribution is accounted for as a positive value, we're able to find that the total area is precisely \(2\). This comprehensive approach guarantees that the spatial area is effectively quantified for practical applications.