Problem 11 - Calculus Early Transcendentals: Pearson New International Edition

Question

use the Second Fundamental Theorem of Calculus to evaluate each definite integral. $$ \int_{0}^{\pi / 2} \cos x d x $$

Step-by-Step Solution

Answer

The integral evaluates to 1.

Step 1: Recall the Second Fundamental Theorem of Calculus

The Second Fundamental Theorem of Calculus states that if a function is continuous on \[ [a, b] \], and \( F \) is an antiderivative of \( f \) on \( [a, b] \), then \[ \int_{a}^{b} f(x) \, dx = F(b) - F(a). \] This tells us we need to find an antiderivative \( F(x) \) for \( \cos x \).

Step 2: Determine the Antiderivative

The antiderivative of \( \cos x \) is \( \sin x \). This means \( F(x) = \sin x \) is an antiderivative of \( f(x) = \cos x \).

Step 3: Evaluate the Antiderivative at the Bounds

Using the antiderivative \( F(x) = \sin x \), we evaluate it at the upper bound \( \pi/2 \) and at the lower bound \( 0 \): \[ F\left(\frac{\pi}{2}\right) = \sin\left(\frac{\pi}{2}\right) = 1 \] and \[ F(0) = \sin(0) = 0. \]

Step 4: Apply the Second Fundamental Theorem of Calculus

Substitute the evaluations into the theorem: \[ \int_{0}^{\pi / 2} \cos x \, dx = F\left(\frac{\pi}{2}\right) - F(0) = 1 - 0 = 1. \] Thus, the value of the definite integral is \( 1 \).

Key Concepts

Definite Integral

A definite integral is a fundamental concept in calculus that represents the exact area under the curve of a function between two points. In essence, it provides a way to accumulate quantities over an interval and amounts to finding the net change over that period.

• Definite integrals are denoted by \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) are the lower and upper limits of integration, respectively.
• The function \( f(x) \) is called the integrand, which is the function being integrated.
• Definite integrals can represent physical concepts such as displacement when integrating velocity or total accumulation when integrating a rate of change.

It's important to note that the definite integral of a function over an interval is always a number, as opposed to an indefinite integral which results in a family of functions. The Second Fundamental Theorem of Calculus helps evaluate these by linking the concept of derivative and integral through antiderivatives.

Antiderivative

An antiderivative is essentially a "backward" derivative, referring to a function \( F(x) \) whose derivative is the given function \( f(x) \). In terms of integration, finding an antiderivative is a crucial step for evaluating a definite integral.

• If \( F'(x) = f(x) \), then \( F(x) \) is an antiderivative of \( f(x) \).
• To find an antiderivative, you often need to think about which function's derivative results in the given integrand.
• In the problem, the antiderivative of \( \cos x \) is \( \sin x \). This is because the derivative of \( \sin x \) is \( \cos x \).

This process allows us to utilize the Second Fundamental Theorem of Calculus to compute definite integrals. By identifying an antiderivative, we can replace the complex problem of integration with simple subtraction of the function values at the given bounds.

Cosine Function

The cosine function, denoted as \( \cos x \), is one of the fundamental trigonometric functions. It describes the horizontal coordinate of a point on the unit circle when swept or rotated through an angle \( x \).

• The cosine function is periodic with a period of \( 2\pi \), meaning it repeats its values every \( 2\pi \) units.
• It is used extensively in trigonometry, with applications spanning from triangles to waves.
• In calculus, the cosine function's integration and differentiation have predictable patterns, such as \( \frac{d}{dx} \cos x = -\sin x \), and its integral is \( \sin x + C \), where \( C \) is a constant of integration for indefinite cases.

Understanding these properties of the cosine function allows you to solve integrals involving it, such as finding its antiderivative which is \( \sin x \). This insight leads to correctly applying the antiderivative in the definite integral challenge presented.