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Chapter 5: Problem 17

Evaluate the integrals in Exercises \(1-28\). $$\int_{0}^{\pi / 8} \sin 2 x d x$$

Short Answer

Expert verified
\( \frac{1}{2} - \frac{\sqrt{2}}{4} \)

Step by step solution

01

Identify the integral to solve

We need to evaluate the integral \( \int_{0}^{\pi / 8} \sin 2x \, dx \). This is a definite integral with the function \( \sin 2x \) and the limits of integration 0 and \( \pi/8 \).
02

Use the substitution method

To solve the integral, we'll use the substitution method. Let \( u = 2x \). Then \( du = 2 \, dx \) or \( dx = \frac{1}{2} \, du \). We will also need to change the limits of integration: when \( x = 0 \), \( u = 0 \); and when \( x = \frac{\pi}{8} \), \( u = \frac{\pi}{4} \).
03

Rewrite the integral with substitution

The integral becomes \( \int_{0}^{\pi/4} \sin u \cdot \frac{1}{2} \, du \). This simplifies to \( \frac{1}{2} \int_{0}^{\pi/4} \sin u \, du \).
04

Integrate \( \sin u \)

The integral of \( \sin u \) is \( -\cos u \). So, we have \( \frac{1}{2} \left[ -\cos u \right]_{0}^{\pi/4} \).
05

Apply the fundamental theorem of calculus

Evaluate the antiderivative from \( u = 0 \) to \( u = \frac{\pi}{4} \):\[ \frac{1}{2} \left( -\cos \frac{\pi}{4} + \cos 0 \right) \].
06

Calculate cosine values

Find the cosine values:- \( \cos \frac{\pi}{4} = \frac{\sqrt{2}}{2} \)- \( \cos 0 = 1 \).Therefore, the expression becomes \( \frac{1}{2} \left( -\frac{\sqrt{2}}{2} + 1 \right) \).
07

Simplify the expression

Simplify \( \frac{1}{2} \left( 1 - \frac{\sqrt{2}}{2} \right) \) to get:\[ \frac{1}{2} \times \left( 1 - \frac{\sqrt{2}}{2} \right) = \frac{1}{2} - \frac{\sqrt{2}}{4} \].

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Key Concepts

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

Substitution Method
The substitution method is a powerful technique used to simplify the process of integration by transforming the original integral into a simpler form. This is achieved by substituting a part of the integral with a new variable, making it easier to solve.
  • Identify a suitable substitution that simplifies the integral. For instance, in our example, we let \( u = 2x \), which transforms the integrand \( \sin 2x \) into \( \sin u \).
  • Calculate the differential of the new variable in terms of the original variable. Here, \( du = 2 \, dx \), leading to \( dx = \frac{1}{2} \, du \).
  • Change the limits of integration if it's a definite integral. Convert \( x \)-limits to \( u \)-limits. When \( x = 0 \), \( u = 0 \) and when \( x = \frac{\pi}{8} \), \( u = \frac{\pi}{4} \).
Applying the substitution method reduces complex trigonometric integrals to simpler forms, allowing them to be easily integrated.
Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus bridges the concept of differentiation and integration, revealing their deep connection. It provides a method for evaluating definite integrals by using antiderivatives.
  • The theorem states that if a function \( F \) is an antiderivative of \( f \) on an interval \([a, b]\), then \( \int_{a}^{b} f(x) \, dx = F(b) - F(a) \).
  • In our solved problem, the antiderivative of \( \sin u \) is \( -\cos u \), allowing us to easily compute the definite integral by evaluating \( -\cos u \) at the upper and lower bounds \( u = \frac{\pi}{4} \) and \( u = 0 \), respectively.
This theorem simplifies the computation of definite integrals, turning them into an algebraic process once the antiderivative is found and evaluated.
Trigonometric Integrals
Trigonometric integrals involve integrating functions that contain trigonometric functions, such as sine or cosine. They often require special techniques to solve, such as substitution, to manage their complexity.
  • In our problem, we integrated \( \sin 2x \) after transforming it into \( \sin u \) using the substitution method.
  • The integral of \( \sin u \) is a standard one: \( -\cos u \). Knowing standard forms of trigonometric integrals is crucial for solving more complicated problems quickly and accurately.
  • Applying limits of integration after finding the antiderivative yields the solution to the original definite integral.
Learning these techniques for handling trigonometric integrals opens the door to solving a wide variety of integration problems involving trigonometric functions.

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Most popular questions from this chapter

Integrals of nonnegative functions Use the Max-Min Inequality to show that if \(f\) is integrable then $$ f(x) \geq 0 \quad \text { on } \quad[a, b] \quad \Rightarrow \quad \int_{a}^{b} f(x) d x \geq 0 $$

The marginal cost of manufacturing \(x\) units of an electronic device is \(0.001 x^{2}-0.5 x+115 .\) If 600 units are produced, what is the production cost per unit?

If your CAS can draw rectangles associated with Riemann sums, use it to draw rectangles associated with Riemann sums that converge to the integrals in Exercises \(89-94 .\) Use \(n=4,10,20,\) and 50 subintervals of equal length in each case. $$ \int_{1}^{2} \frac{1}{x} d x \quad(\text { The integral's value is about } 0.693 .) $$

Revenue from marginal revenue Suppose that a company's marginal revenue from the manufacture and sale of eggbeaters is $$\frac{d r}{d x}=2-2 /(x+1)^{2},$$ where \(r\) is measured in thousands of dollars and \(x\) in thousands of units. How much money should the company expect from a production run of \(x=3\) thousand eggbeaters? To find out, integrate the marginal revenue from \(x=0\) to \(x=3\) .

In Exercises \(47-50,\) find the total area between the region and the \(x\) -axis. $$y=3 x^{2}-3, \quad-2 \leq x \leq 2$$
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