Question

Suppose \(p\) is a nonzero real number and \(f\) is an odd function with \(\int_{0}^{1} f(x) d x=\pi .\) Evaluate each integral. a. \(\int_{0}^{\pi /(2 p)}(\cos p x) f(\sin p x) d x\) b. \(\int_{-\pi / 2}^{\pi / 2}(\cos x) f(\sin x) d x\)

Short answer

1. \(\int_{0}^{\pi /(2p)} (\cos px) f(\sin px) dx\) 2. \(\int_{-\pi / 2}^{\pi / 2}(\cos x) f(\sin x) dx\) Answer: 1. The value of the first integral is \(\frac{\pi}{p}\). 2. The value of the second integral is \(0\).

Step-by-step solution

Write down given integral

First, let's write down the given integral: \(\int_{0}^{\pi /(2p)}(\cos px) f(\sin px) dx\)

Apply substitution

Let's apply substitution: \(u = \sin px\). Then, \(du = p\cos px dx\). This yields the new bounds for the integral: when \(x = 0\), \(u = \sin(0) = 0\), and when \(x = \frac{\pi}{2p}\), \(u = \sin\frac{\pi}{2} = 1\), so: \(\int_{0}^{\pi /(2 p)}(\cos p x) f(\sin p x) d x = \int_0^1 \frac{1}{p} f(u) du\)

Multiply constant outside the integral

Now, we move the constant \(\frac{1}{p}\) out of the integral: \(\int_0^1 \frac{1}{p} f(u) du = \frac{1}{p} \int_0^1 f(u) du\)

Final expression using given relationship

We know that \(\int_{0}^{1} f(x) dx = \pi\). So, replacing \(u\) with \(x\) in the integral and using the given relationship: \(\frac{1}{p} \int_0^1 f(x) dx = \frac{1}{p}\pi\) Hence, the value of the first integral is \(\frac{\pi}{p}\). For the second integral:

Write down given integral

Let's write down the given integral: \(\int_{-\pi / 2}^{\pi / 2}(\cos x) f(\sin x) dx\)

Apply substitution

Again, we apply substitution: \(u = \sin x\). Then, \(du = \cos x dx\). This yields the new bounds for the integral: when \(x = -\frac{\pi}{2}\), \(u = -1\), and when \(x = \frac{\pi}{2}\), \(u = 1\), so: \(\int_{-\pi / 2}^{\pi / 2}(\cos x) f(\sin x) dx = \int_{-1}^1 f(u) du\)

Use odd function property

Since \(f(x)\) is an odd function, we have \(\int_{-a}^a f(u) du = 0\) for any \(a\). In this case, we have: \(\int_{-1}^1 f(u) du = 0\) The value of the second integral is \(0\).

Key concepts

Odd Functions

Odd functions are a central concept in calculus, particularly when working with definite integrals. An odd function is defined by the property that for every input, the function's output at the negative of that input is the negative of the output at the input. Mathematically, this is expressed as:

• If a function is odd, then for all values of the variable, we have:

\[f(-x) = -f(x)\] This property has significant implications when evaluating definite integrals. Specifically,

• If you integrate an odd function from \( -a \) to \( a \), the result is always zero:
• \[ \int_{-a}^{a} f(x) \, dx = 0 \]

This occurs because the contributions from the negative and positive subintervals cancel each other out. In essence, the areas under the curve on either side of the y-axis are equal in magnitude but opposite in sign. For example, in our exercise with the function \( f(x) \), this property was used to deduce that:

• \(\int_{-1}^1 f(u) du = 0\)

Understanding this property simplifies many integral calculations and is particularly useful in symmetrically defined intervals.

Substitution Method

The substitution method is a powerful technique for solving integrals, especially when dealing with complicated functions that involve composition, such as when the function inside the integral is a product of several functions. This technique is akin to reversing the chain rule from differentiation. Here's how it works:

• First, choose a substitution that simplifies the integral. In our exercise, the substitution \( u = \sin(px) \) was used.
• Once the substitution variable is chosen, find its derivative and rewrite the integral in terms of \( u \):\( du = p\cos(px) \, dx \).
• With the substitution, adjust the limits of integration. When \(x = 0\), \(u = \sin(0) = 0\); when \(x = \frac{\pi}{2p}\), \(u = 1\).
• Transform the integral using this substitution, which results in \( \int_0^1 \frac{1}{p} f(u) \, du \). Notice how the function has become simpler.
• Integrate with respect to \( u \), and once complete, substitute back to the original variable if required.

This method not only aids in tackling integrals that initially appear complex, but it also can transform difficult boundaries into simpler ones, making the integral more manageable.

Definite Integrals

Definite integrals are a fundamental concept in calculus that extend the idea of a sum of infinitesimally small quantities across a continuous interval. Unlike indefinite integrals, which result in a family of functions (anti-derivatives), definite integrals yield a specific numerical value. Here's what makes definite integrals special:

• They are written as \( \int_a^b f(x) \, dx \), where \( a \) and \( b \) are the limits of integration that define the interval over which the function is integrated.
• The result of a definite integral represents the net area under the curve \( f(x) \) from \( x = a \) to \( x = b \), considering above and below the x-axis.
• In the case of the given exercise, definite integrals were used to evaluate the function \( f(x) \) with specific intervals.
• Important properties often utilized include the linearity of integrals, such as the ability to factor out constants, demonstrated as \( \frac{1}{p} \int_0^1 f(u) \, du \).

Overall, definite integrals are instrumental in solving a variety of practical problems in physics, engineering, and statistics. They measure quantities such as area, displacement, and total accumulation.