Question

if \(\mathrm{f}(\mathrm{x})=\mathrm{x} \int_{0} \log [(1-\mathrm{t}) /(1+\mathrm{t})] \mathrm{dt}\) then \(\mathrm{f}(1 / 2)-\mathrm{f}[(-1) / 2]\) is equals (a) 0 (b) \((1 / 2)\) (c) \([(-1) / 2]\) (d) 1

Short answer

The short answer is: \(f(\frac{1}{2}) - f(\frac{-1}{2}) = 0\) The correct option is (a) 0.

Step-by-step solution

Find the closed form of the integral

To find the closed form of the given integral, we can start by expanding the logarithm using log(a/b) = log(a) - log(b), then integrate the resulting expression with respect to t. The given integral is: \(\int_{0} \log[(1-t)/(1+t)] dt\) Expanding the logarithm, we have: \(\int_{0} (\log(1-t) - \log(1+t)) dt\) Now, we will find the antiderivative of each term separately. For the first term, let u = 1 - t, then du = -dt, and dt = -du. So we get: \(-\int_{0} \log(u) du\) We know that the antiderivative of -log(u) is u * (1 + log(u)). Now, we'll find the antiderivative of the second term: For the second term, let v = 1 + t, then dv = dt, and dt = dv. \(\int_{0} \log(v) dv\) We know that the antiderivative of log(v) is v * (-1 + log(v)). Now, we can write the closed form of the integral: \(\int_{0} \log[(1-t)/(1+t)] dt = u * (1 + log(u)) - v * (-1 + log(v))\) .MapFrom(u,v)->(1-t,1+t) \((1-t)(1+\log(1-t))-(1+t)(-1+\log(1+t))\)

Evaluate f(x) for x = 1/2 and x = -1/2

We have found that: \(f(x) = x[(1 - x)(1 + \log(1 - x)) - (1 + x)(-1 + \log(1 + x))]\) Let's evaluate f(x) for x = 1/2: \(f(1/2) = (\frac{1}{2})[((1 - \frac{1}{2})(1 + \log(1 - \frac{1}{2}))-(1 + \frac{1}{2})(- 1 + \log(1 + \frac{1}{2}))]\) \(f(1/2)=\frac{1}{2}(0)\) Now, let's evaluate f(x) for x = -1/2: \(f(-1/2) = (\frac{-1}{2})[((1 - \frac{-1}{2})(1 + \log(1 - \frac{-1}{2}))-(1 + \frac{-1}{2})(- 1 + \log(1 + \frac{-1}{2}))]\) \(f(-1/2)=\frac{-1}{2}(0)\)

Calculate f(1/2) - f(-1/2) and find the correct option

Now, we can calculate the difference between f(1/2) and f(-1/2): \(f(1/2) - f(-1/2) = \frac{1}{2}(0) - \frac{-1}{2}(0) = 0\) Comparing this result with the given options, we can see that: \(f(1/2) - f(-1/2) = 0\) This matches option (a). Therefore, the correct answer is: (a) 0

Key concepts

Integration Techniques

Integration techniques are crucial for solving complex integrals, especially when direct integration is not possible. One common technique involves simplifying the integral by using properties of logarithms.

• Consider the identity: \[ \log\left(\frac{a}{b}\right) = \log(a) - \log(b) \] This allows you to separate complex expressions into simpler components that are easier to integrate.
• In this exercise, we applied this property to separate the integral into two separate integrals, allowing us to find the antiderivatives.
• An important step in integration is the change of variables, where we set new variables for each separated term, thus simplifying integration.

Understanding these techniques enables you to tackle integrals that at first seem unsolvable, transforming them into more manageable forms.

Antiderivatives

Antiderivatives, also known as indefinite integrals, represent the reverse process of differentiation. When we find an antiderivative of a function, we're essentially asking: what function, when differentiated, gives us the original function?

• In the context of this problem, after separating the logarithmic terms, \(\log(1-t)\) and \(\log(1+t)\), we found their respective antiderivatives.
• For a term like \(\log(u)\), where \(u\) is a transformed variable, the antiderivative is: \[ u \cdot (1 + \log(u)) \]
• It's vital to evaluate the antiderivative across the specified limits after reintegrating, as these determine the value of a definite integral.

Finding antiderivatives requires a blend of formula knowledge and practice, enhancing your capability to solve diverse integral equations.

Logarithmic Functions

Logarithmic functions are inverse operations of exponential functions and play a significant role in integration. When dealing with integrals involving logarithms, understanding their properties and identities is vital.

• The key property used in this exercise is the decomposition of a logarithmic fraction: \[ \log \left( \frac{1-t}{1+t} \right) = \log(1-t) - \log(1+t) \]
• Such transformations allow us to break down complex integrals into simpler parts that are easier to handle analytically.
• This knowledge is fundamental not only in integration exercises but also in various fields such as engineering, physics, and computer science, where logarithmic expressions frequently appear.

Mastering logarithmic functions and their properties is critical for simplifying problems and finding solutions effectively.