Problem 59 - Calculus Early Transcendentals

Question

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

Step-by-Step Solution

Answer

Question: Find the area of the region bounded by the graph of \(f(x) = \sin x\) and the x-axis on the interval \([-\pi/4, 3\pi/4]\). Answer: The area of the region enclosed by the graph of \(f\) and the x-axis on the interval \([-\pi/4, 3\pi/4]\) is \(-\frac{1}{\sqrt{2}} +\frac{1}{\sqrt{2}}\).

Step 1: Set up the definite integral

We want to find the area enclosed by the function and the x-axis on the interval \([-\pi/4, 3\pi/4]\). To do this, we will compute the definite integral of the function \(\sin x\) from \(-\pi/4\) to \(3\pi/4\). This can be represented as: $$ \int_{-\pi / 4}^{3\pi / 4} \sin x\,dx $$

Step 2: Compute the antiderivative of the function

To compute the definite integral, we first need to find the antiderivative of the function \(\sin x\). The antiderivative of \(\sin x\) is given by the function \(-\cos x\).

Step 3: Evaluate the antiderivative at the endpoints of the interval

Now that we have the antiderivative of the function, we can plug in the endpoints of the interval \([-\pi/4, 3\pi/4]\) to evaluate the definite integral. We have: $$ \begin{aligned} -\cos(3\pi / 4) - (- \cos(-\pi / 4)) &= -\frac{1}{\sqrt{2}} - ( -\frac{1}{\sqrt{2}} ) \\ &= -\frac{1}{\sqrt{2}} +\frac{1}{\sqrt{2}} \\ \end{aligned}$$

Step 4: Find the area of the region enclosed by the function and the x-axis

Now that we have the definite integral, we can find the area of the region enclosed by the graph of \(f\) and the x-axis on the given interval. We just need to plug back the evaluated antiderivative to find the area: $$ \begin{aligned} \int_{-\pi / 4}^{3\pi / 4} \sin x\,dx &= \left[-\cos(3\pi / 4) - (- \cos(-\pi / 4))\right] \\ &= -\frac{1}{\sqrt{2}} +\frac{1}{\sqrt{2}} \\ \end{aligned}$$ So the area of the region enclosed by the graph of \(f\) and the x-axis on the interval \([-\pi/4, 3\pi/4]\) is \(-\frac{1}{\sqrt{2}} +\frac{1}{\sqrt{2}}\).

Key Concepts

Antiderivative

The antiderivative, also known as an indefinite integral, is essentially the reverse process of differentiation. It's about finding a function whose derivative is the given function. In our problem, we have the function \( f(x) = \sin x \). To find its antiderivative, we need a function that, when differentiated, returns \( \sin x \). This function is \( -\cos x \), because the derivative of \( -\cos x \) is \( \sin x \). This step is crucial when solving definite integrals since it allows us to evaluate the area under a curve over a specific interval.

Trigonometric Functions

Trigonometric functions, such as sine, cosine, and tangent, are fundamental in mathematics and come up often in calculus. Here, we're dealing with the sine function, \( \sin x \), which is periodic and oscillates between -1 and 1. When you visualize \( \sin x \) on a graph, it creates a wave-like pattern. Understanding how these functions behave is important for finding areas and solving equations. For instance, knowing that the antiderivative of \( \sin x \) is \( -\cos x \) helps us compute the area in this problem. Working with these functions often involves angles measured in radians, which is why you see terms like \( \pi/4 \).

Area Under Curve

Finding the area under a curve is a common application of definite integrals. It involves calculating the integral of the function over a specified interval. In our exercise, we look at the integral from \(-\pi/4\) to \(3\pi/4\) of \( \sin x \). This specific definite integral tells us the net area between the curve and the x-axis from the starting point to the endpoint. By evaluating the antiderivative, \( -\cos x \), at the interval's endpoints and subtracting the results, the enclosed area is found. Although our final answer here turns out to be zero, which indicates symmetrical areas canceling each other out, this process illustrates how definite integrals allow us to determine such areas accurately.