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Chapter 16: Problem 22

Evaluate the integrals. \(\int_{\pi / 2}^{\pi} \cos x d x\)

Short Answer

Expert verified
The integral of the function \(\cos{x}\) from \(\pi/2\) to \(\pi\) is -1.

Step by step solution

01

Find the antiderivative

The antiderivative of the function \(f(x) =\cos(x)\) is: \[ F(x) = \int \cos(x) dx = \sin(x) + C \] Here, \(F(x)\) is the antiderivative and \(C\) represents the constant of integration. However, since we are evaluating a definite integral, we don't need to worry about the constant of integration. Thus, we can ignore \(C\) in this case.
02

Use the Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus states that if a function is continuous on the interval \([a, b]\) and has an antiderivative \(F(x)\), then: \[ \int_{a}^{b} f(x) dx = F(b) - F(a) \] In our case, the function \(\cos{x}\) is continuous on the interval \([\pi/2, \pi]\). We previously found the antiderivative to be \(F(x) = \sin(x)\). Now, we will plug our limits and antiderivative into the formula:
03

Evaluate the definite integral

Using the Fundamental Theorem of Calculus, we have: \[ \int_{\pi/2}^{\pi} \cos(x) dx = \sin(\pi) - \sin(\pi/2) \] Now, we evaluate \(sin(\pi)\) and \(sin(\pi/2)\): \[ \sin(\pi) = 0 \qquad \textrm{and} \qquad \sin(\pi/2) = 1 \] So, the definite integral becomes: \[ \int_{\pi/2 }^{\pi} \cos(x) dx = \sin(\pi) - \sin(\pi/2) = 0 - 1 = -1 \] Therefore, the integral of the function \(\cos{x}\) from \(\pi/2\) to \(\pi\) is -1.

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Key Concepts

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

Antiderivative
Finding an antiderivative is like discovering the reverse of differentiation. When we differentiate a function, we find its derivative. However, the antiderivative is the original function from which a given function (derivative) is derived.
In the context of a definite integral, we use this antiderivative without worrying about the constant that usually comes with indefinite integrals. For the function \(f(x) = \cos(x)\), the antiderivative is \(F(x) = \sin(x)\). This is because when you differentiate \(\sin(x)\), you get \(\cos(x)\).
  • Important Note: In definite integrals, we ignore the constant of integration \(C\). This constant cancels out when subtracting \(F(a)\) from \(F(b)\).
Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus (FTC) bridges differentiation and integration, two main operations in calculus. It offers a way to find the area under a curve easily using antiderivatives. The theorem states:
If a function is continuous on an interval \([a, b]\) and \(F\) is an antiderivative of \(f\) within that interval, then:\[\int_{a}^{b} f(x) dx = F(b) - F(a)\]This mighty statement enables us to evaluate definite integrals without the tedious task of limit calculations.
  • Example: For \(\int_{\pi/2}^{\pi} \cos(x) dx\), using \(F(x) = \sin(x)\), we find the value by calculating \(\sin(\pi) - \sin(\pi/2) = 0 - 1 = -1\).
Continuous Function
A function is considered continuous over an interval if there are no breaks, holes, or jumps within that interval. This means you can draw it without lifting your pen off the paper.
The continuity of a function is a crucial requirement for the Fundamental Theorem of Calculus to apply. In our exercise, \(\cos(x)\) is a continuous function on the interval \([\pi/2, \pi]\), making it suitable for evaluating the definite integral.
  • Key Features of Continuous Functions:
    • No interruptions in the graph!
    • The limit of the function at any point in its domain equals the function's value at that point.

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