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Chapter 4: Problem 107

Prove that $$ \int_{-1 / 2}^{1 / 2} 2^{\cos x} d x=2 \int_{0}^{1 / 2} 2^{\cos x} d x $$

Short Answer

Expert verified
The function \(2^{\cos x}\) is even, as \(f(-x) = 2^{\cos(-x)} = 2^{\cos x} = f(x)\). Using the property of even functions, \(\int_{-a}^{a} f(x) dx = 2\int_{0}^{a} f(x) dx\), we can simplify the given integral: \(\int_{-1/2}^{1/2} 2^{\cos x} dx = 2\int_{0}^{1/2} 2^{\cos x} dx\) Since both sides of the equation are identical, we have proven the given equation to be true.

Step by step solution

01

Identify that the function 2^(\cos x) is even

An even function is a function where f(-x) = f(x) for all x in its domain. Let's check if the function 2^{\cos x} is even: \[ f(-x) = 2^{\cos(-x)} = 2^{\cos x} = f(x) \] Since the function satisfies the condition for an even function, we can proceed to the next step.
02

Use the property of even functions to simplify the left-hand side of the given equation

The property of even functions states that for any even function f(x), the following holds: \[ \int_{-a}^{a} f(x) dx = 2\int_{0}^{a} f(x) dx \] In our case, the function f(x) is 2^{\cos x} and the interval is \([-\frac{1}{2},\frac{1}{2}]\). Therefore, we can rewrite the left-hand side of the given equation: \[ \int_{-1/2}^{1/2} 2^{\cos x} dx = 2\int_{0}^{1/2} 2^{\cos x} dx \]
03

Compare the simplified equation to the right-hand side of the given equation

We have shown that the left-hand side of the given equation is equal to the property of even functions with the function \(2^{\cos x}\) and the interval \([-\frac{1}{2},\frac{1}{2}]\): \[ \int_{-1/2}^{1/2} 2^{\cos x} dx = 2\int_{0}^{1/2} 2^{\cos x} dx \] Now we see that the right-hand side of the given equation is exactly the same: \[ 2\int_{0}^{1/2} 2^{\cos x} dx \] Since both sides of the equation are identical, we have proven the given equation to be true.

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Key Concepts

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

Even Functions
Even functions have a special property where if you take the function, say \( f(x) \), and replace \( x \) with \( -x \), the output remains unchanged. This means that \( f(-x) = f(x) \). To think of it visually, an even function is symmetrical about the y-axis.

Consider the example from the exercise, where \( f(x) = 2^{\cos x} \). We need to verify if it is an even function. Start by calculating \( f(-x) \):
  • \( f(-x) = 2^{\cos(-x)} \)
  • Since \( \cos(-x) = \cos x \) (a trigonometric identity), this gives \( f(-x) = 2^{\cos x} = f(x) \)
Thus, \( f(x) \) satisfies the condition for an even function. This symmetry property is useful because it helps in evaluating integrals over symmetric intervals.
Definite Integrals
Definite integrals provide a way to calculate the accumulation of quantities over a specific interval. The expression \( \int_{a}^{b} f(x) \, dx \) represents the area under the curve \( f(x) \) from \( x = a \) to \( x = b \).

In the context of our exercise, we have an integral:
  • \( \int_{-1/2}^{1/2} 2^{\cos x} \, dx \)
This integral calculates the total accumulation of the function \( 2^{\cos x} \) over the interval from \(-\frac{1}{2}\) to \(\frac{1}{2}\).

The role of definite integrals is crucial as they are used to determine areas, volumes, and in this case, verify certain symmetries and properties of functions.
Properties of Integrals
Integral calculus comes with a rich set of properties that simplify calculations. For even functions, one important property can greatly help:
  • \( \int_{-a}^{a} f(x) \, dx = 2 \int_{0}^{a} f(x) \, dx \) for an even function \( f(x) \)
This property implies that when you calculate the definite integral of an even function over a symmetric interval centered at zero, you only need to calculate the integral from zero to \( a \) and double it.

For the example in the exercise, the property is applied to the function \( 2^{\cos x} \) over the interval \([- rac{1}{2}, rac{1}{2}]\), making it simpler to evaluate:
  • Subtract the need to calculate separate areas under positive and negative x values.
  • Saves time by reducing the complexity of integration.
Understanding these properties not only simplifies solving problems but also enhances comprehension of integral calculus as a tool for mathematical analysis.

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Most popular questions from this chapter

Determine whether the statement is true or false. If it is true, explain why it is true. If it is false, explain why or give an example to show why it is false. $$ \int f^{\prime}(x) d x=f(x)+C $$

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Air Pollution According to the South Coast Air Quality Management District, the level of nitrogen dioxide, a brown gas that impairs breathing, present in the atmosphere on a certain June day in downtown Los Angeles is approximated by $$A(t)=0.03 t^{3}(t-7)^{4}+62.7 \quad 0 \leq t \leq 7$$ where \(A(t)\) is measured in pollutant standard index and \(t\) is measured in hours with \(t=0\) corresponding to 7 A.M. What is the average level of nitrogen dioxide present in the atmosphere from 7 A.M. to 2 P.M. on that day?

The percentage of a current Mediterranean population with serum cholesterol levels between 160 and \(180 \mathrm{mg} / \mathrm{dL}\) is estimated to be $$P=\sqrt{\frac{2}{\pi}} \int_{160}^{180} e^{-(1 / 2)[(x-160) / 50]^{2}} d x$$ Estimate \(P\).

Find the indefinite integral. $$ \int x^{2}(1-x)^{7} d x $$
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