Question

Consider the function \(f(x)=\frac{1-x}{x}\). a. Are there numbers \(01\) such that \(\int_{1 / a}^{a} f(x) d x=0 ?\)

Short answer

Question: Determine if there are numbers (a) in the range of \(01\) such that the given conditions are satisfied. Solution: After calculating the indefinite integral of the given function \(f(x)=\frac{1-x}{x}\), we found its antiderivative to be \(\ln|x| - x + C\). For the range \(01\), we evaluated the definite integral and obtained the equation: $$-a+\ln (a^{2}) -\frac{1}{a} + 1=0$$ Similarly, using a graphical approach, it can be determined whether or not a solution exists for \(a>1\).

Step-by-step solution

Calculate the indefinite integral of f(x)

The given function is \(f(x)=\frac{1-x}{x}\). To find its antiderivative, we want to compute the integral: $$\int f(x) dx=\int \frac{1-x}{x} dx$$ We can split the fraction into two separate fractions: $$\int (\frac{1}{x} - 1) dx$$ Now, we can integrate each term separately: $$\int \frac{1}{x} dx - \int 1 dx = \ln|x| - x + C$$ Here C is the constant of integration.

Step 2a: Evaluate the definite integral for the range \(0

We want to find out if there are numbers \(0<a<1\) such that: $$\int_{1-a}^{1+a} f(x) dx=0$$ Using the obtained antiderivative, we calculate the definite integral as follows: $$\left[\ln|x| - x\right]_{1-a}^{1+a} = \left(\ln(1+a) -(1+a)\right) - \left(\ln(1-a) -(1-a)\right)$$ To check whether the integral equals zero, we must simplify the expression and see if we can find any \(a\) in the specified range: $$\ln\frac{1+a}{1-a} -2a = 0$$ Since finding an explicit solution for \(0<a<1\) might be difficult, we can use a graphical approach to see if this equation has a solution in this range. We can plot the function $$y=\ln\frac{1+x}{1-x}-2x$$ for \(0<x<1\). If the function crosses the x-axis in this range, it indicates that a solution exists for \(0<a<1\).

Step 2b: Evaluate the definite integral for the range \(a>1\)

We want to find out if there are numbers \(a>1\) such that: $$\int_{1 / a}^{a} f(x) dx=0$$ Using our found antiderivative, we can evaluate the definite integral as follows: $$\left[\ln|x| - x\right]_{1/a}^{a} =\left(\ln a -a\right) -\left(\ln\frac{1}{a} -\frac{1}{a}\right)$$ which simplifies to: $$-a+\ln (a^{2}) -\frac{1}{a} + 1=0$$ Again, finding an explicit solution for the range \(a>1\) might be difficult, but we can use the same graphical approach as in the previous step. We can plot the function $$y=-x+\ln (x^{2}) -\frac{1}{x} + 1$$ for \(x>1\). If the function crosses the x-axis in this range, it indicates that a solution exists for \(a>1\).

Key concepts

Indefinite Integral

An indefinite integral, often referred to as an antiderivative, is a fancy term for finding a function that reverses differentiation. When you integrate a function without limits, you're essentially uncovering all the functions whose derivative is the original function. The result includes a constant of integration (denoted as ‘C’) because differentiation of a constant yields zero, meaning there could be infinite antiderivatives!
To illustrate, given a function like \(f(x) = \frac{1-x}{x}\), the indefinite integral tells us the general form of the antiderivative:

• Begin by rewriting the function in a simpler form: \(f(x) = \frac{1}{x} - 1\).
• The indefinite integral is then \(\int (\frac{1}{x} - 1) \; dx\).
• Integrate each term separately: \(\int \frac{1}{x} \; dx = \ln|x| + C_1\) and \(\int -1 \; dx = -x + C_2\).
• Combine these results to get the antiderivative: \(\ln|x| - x + C\), where \(C\) represents any constant value.

This process gives the general solution, teaching us not just numbers, but functions that reveal broader patterns in calculus.

Antiderivative

The antiderivative is essentially the reverse process of differentiation. If you take the derivative of an antiderivative, you land back at your original function. This unique characteristic is what ties together with indefinite integrals. In the exercise, finding the antiderivative is critical for evaluating definite integrals later.
For our function \(f(x) = \frac{1-x}{x}\), the antiderivative is found by integrating term-by-term to get \(\ln|x| - x + C\). Thus, the antiderivative reveals all the functions that could have the same slope as \(f(x)\) at every point. It is quite versatile:

• Any arithmetic operation, when reversed, should point back to the original function, showing how crucial basics are in uncovering deeper truths in calculus.
• In solving problems, understanding antiderivatives enables us to assess changes across intervals, which is exactly why we link antiderivatives to definite integrals next.

Recognizing the antiderivative not just revisits familiar territory of differentiation but enhances understanding to solve complex integrals like in the original task.

Integration by Parts

Integration by Parts is a powerful technique for solving integrals that can't be solved by simple manipulation. It's based on the product rule of differentiation and is particularly useful when dealing with products of functions. Though not directly applied in the original exercise, understanding it is valuable for tackling complex integrals.
The formula for integration by parts is derived from the product rule and is stated as: \[\int u \; dv = uv - \int v \; du\]
Here's how to apply it:

• Identify parts of the integrand. Let one part be \(u\) and the differential part \(dv\).
• Compute \(du\) by differentiating \(u\) and find \(v\) by integrating \(dv\).
• Substitute known values into the formula to simplify the integral.
• This transformation often simplifies solving otherwise stubborn integrals.

While this might look daunting at first, practice turns integration by parts into a handy tool in your calculus toolkit. By mastering such techniques, you can handle a range of integral problems more effectively, including those you may encounter alongside the exercise provided.