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Chapter 5: Problem 94

Working with area functions Consider the function \(f\) and the points \(a, b,\) and \(c\) a. Find the area function \(A(x)=\int_{a}^{x} f(t) d t\) using the Fundamental Theorem. b. Graph \(f\) and \(A\) c. Evaluate \(A(b)\) and \(A(c) .\) Interpret the results using the graphs of part \((b)\) $$f(x)=\sin x ; a=0, b=\frac{\pi}{2}, c=\pi$$

Short Answer

Expert verified
Answer: At \(x = \frac{\pi}{2}\), the area under the curve of \(f(x) = \sin x\) and the x-axis from \(a=0\) to \(b = \frac{\pi}{2}\) is 1 square unit. At \(x = \pi\), the area under the curve of \(f(x) = \sin x\) and the x-axis from \(a=0\) to \(c = \pi\) is 2 square units.

Step by step solution

01

Find the area function using the Fundamental Theorem

By the Fundamental Theorem, the area function \(A(x)\) is given by the definite integral: $$A(x) = \int_a^x f(t)dt$$. In this case, we have the function \(f(x) = \sin x\) and the lower bound \(a = 0\). The area function becomes: $$A(x) = \int_0^x \sin t dt$$ Now, we need to evaluate the integral to find \(A(x)\).
02

Evaluate the integral

To evaluate the integral, we need to find the antiderivative of the function \(\sin t\). We know that the derivative of \(\cos t\) is \(-\sin t\). Thus, the antiderivative of \(\sin t\) is \(-\cos t\): $$A(x) = \int_0^x \sin t dt = -\cos t \Big|_0^x$$ Now, let's apply the limits of the integral.
03

Apply the limits to the area function

Applying the limits, we get: $$A(x) = -\cos x - (-\cos 0) = -\cos x + 1$$ This is the area function we were looking for.
04

Graph the function and the area function

In this step, we graph the function \(f(x) = \sin x\) and the area function \(A(x) = -\cos x + 1\). For the function \(f(x) = \sin x\), it is a wave-like function having a period of \(2\pi\) with amplitude 1. For the area function \(A(x) = -\cos x + 1\), it is a wave-like function with a period of \(2\pi\) and amplitude 1, but shifted upward by 1 unit.
05

Evaluate A(b) and A(c)

Now we have to find the values of the area function for \(b = \frac{\pi}{2}\) and \(c = \pi\). $$A\left(\frac{\pi}{2}\right) = -\cos\left(\frac{\pi}{2}\right) + 1 = -0 + 1 = 1$$ $$A(\pi) = -\cos(\pi) + 1 = -(-1) + 1 = 2$$
06

Interpret the results using the graphs

Using the graphs of the function \(f(x) = \sin x\) and the area function \(A(x) = -\cos x + 1\), we can interpret the results of part (c). For \(x = \frac{\pi}{2}\), the area under the curve is 1, meaning that the area enclosed between the function \(f(x) = \sin x\) and the x-axis from \(a = 0\) to \(b = \frac{\pi}{2}\) is 1 square unit. For \(x = \pi\), the area under the curve is 2. This implies that the area enclosed between the function \(f(x) = \sin x\), and the x-axis from \(a = 0\) to \(c = \pi\) is 2 square units.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Definite Integral
The definite integral is a fundamental concept in calculus, often used to evaluate the area under a curve. It represents the accumulated value of a function over a specific interval. When dealing with a definite integral, you are essentially looking at the sum of all tiny slices of area that fit under the curve of the function within the given limits. For example, when we integrate the function \(f(x) = \sin x\) from \(0\) to \(x\), it helps us find the area function \(A(x) = \int_0^x \sin t \, dt\). This is achieved by determining the antiderivative of \(\sin x\) and applying the limits of integration, which offers a complete picture of the area accumulated up to that point.
Antiderivative
An antiderivative is a function that "reverses" the process of differentiation. If the derivative of a function \(F(x)\) is \(f(x)\), then \(F(x)\) is an antiderivative of \(f(x)\). In the context of finding the area under a curve, the antiderivative helps simplify the process of evaluating a definite integral. For our specific function \(f(x) = \sin x\), the antiderivative is \(-\cos x\) because the derivative of \(-\cos x\) is \(\sin x\). Thus, using this antiderivative, we compute \(A(x) = -\cos x + 1\) after applying the limits to the integral, giving us the function that describes the area under the curve of \(\sin x\) from 0 to \(x\).
Area Under a Curve
Finding the area under a curve involves determining the space that lies beneath a given function and above the x-axis within a specified interval. This area helps us understand how much space the function encompasses. In our example with \(f(x) = \sin x\), the area function \(A(x) = -\cos x + 1\) is particularly insightful because it quantifies the area underneath \(\sin x\) and above the x-axis from 0 to \(x\). Evaluating this at specific points, such as \(x = \frac{\pi}{2}\) and \(x = \pi\), we find that the area is \(1\) square unit for the former and \(2\) square units for the latter. These results tell us the accumulated area under the sine curve over those intervals.
Trigonometric Functions
Trigonometric functions like \(\sin x\) and \(\cos x\) play critical roles in calculus and many real-world applications. These periodic functions are often used to model wave-like behavior. The function \(f(x) = \sin x\) represents a wave that oscillates between -1 and 1, with a period of \(2\pi\). In the context of our problem, \(\sin x\) is integrated to study the accumulation of area. The antiderivative \(-\cos x\) also follows a similar wave pattern but is shifted. Understanding these functions and their characteristics is essential for interpreting their role in calculating areas, modeling behaviors, and solving various practical problems in mathematics and physics.

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Most popular questions from this chapter

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