Question

If \(a<0

Short answer

The short answer is: \(b\int_a^b\frac{|x|}{x} dx = ab + b^2\)

Step-by-step solution

Rewrite the function with the absolute value function

The given function is: \(|x|/x\) We can rewrite the function without using the absolute value by considering the cases when x is positive or negative. If x < 0, then |x| = -x, and |x|/x = -1 If x > 0, then |x| = x, and |x|/x = 1 Now, we rewrite the function as: \[f(x) = \begin{cases} -1 & \text{if } x < 0 \\ 1 & \text{if } x \geq 0 \end{cases} \]

Break the integral using the cases

Now we will break the integral into two parts using the function definition above and the given bounds (a, b). Since a < 0 and b > 0, the integral can be split at x = 0: \[b\int_a^b\frac{|x|}{x} dx = b\int_a^0\frac{-x}{x} dx + b\int_0^b\frac{x}{x} dx\]

Evaluate the integrals

Now, we evaluate the integrals for each part: \[b\int_a^0\frac{-x}{x} dx = b\int_a^0(-1) dx = -b[x]_a^0 = -b(0-a) = ab\] \[b\int_0^b\frac{x}{x} dx = b\int_0^b 1 dx = b[x]_0^b = b(b-0) = b^2\]

Combine the results and simplify

Now, we combine the results from step 3 and simplify: \(b\int_a^b\frac{|x|}{x} dx = ab + b^2\)

Identify the correct option

Now, we check the given multiple-choice options to find the correct answer: (a) a + b: This cannot be the correct answer since our result contains an extra b term. (b) b - a: This also cannot be the correct answer since our result contains a squared term. (c) a - b: This cannot be the correct answer since our result contains a squared term. (d) \((b - a) / 2\): This also cannot be the correct answer since our result does not divide the expression by 2. None of the options match the result we obtained (ab + b^2). There is a possibility of an error in the given options, so we conclude with our obtained result: \[b\int_a^b\frac{|x|}{x} dx = ab + b^2\]

Key concepts

Piecewise Functions

Piecewise functions are mathematical expressions that have different definitions depending on the input value. They are usually defined by multiple sub-functions, each applicable to a specific interval within the domain. This allows a function to adapt its behavior based on the value of the input variable.
For example, a piecewise function might be defined as follows:

• When the input is less than zero, the function outputs \(f(x) = -1\).
• When the input is greater than or equal to zero, it outputs \(f(x) = 1\).

This is exactly what we have in the case of the function \(f(x) = \frac{|x|}{x}\). It's broken down into two pieces:

• For \(x < 0\), we have \(f(x) = -1\).
• For \(x \geq 0\), we have \(f(x) = 1\).

Such piecewise functions are particularly useful in handling complex scenarios like those involving absolute values.

Absolute Value Function

The absolute value function, denoted by \(|x|\), measures the distance of a number from zero on the number line. It is always non-negative and is defined as follows:

• If \(x \geq 0\), then \(|x| = x\).
• If \(x < 0\), then \(|x| = -x\).

This function effectively removes any negative signs from its input, reflecting negative values to their positive counterparts. In our context, we used the absolute value function to determine how the expression \(\frac{|x|}{x}\) changes sign based on the value of \(x\).
By breaking it down to two cases, we observed that \(\frac{|x|}{x} = -1\) for negative \(x\) and \(+1\) for non-negative \(x\), simplifying our calculations for integration.

Integration by Parts

While integration by parts is not explicitly used in this particular exercise, it's an important technique in calculus for integrating products of functions. This method is derived from the product rule of differentiation, and its formula is:
\[\int u \, dv = uv - \int v \, du\]
Here, \(u\) and \(v\) are differentiable functions, and \(dv\) and \(du\) are their respective derivatives.
The concept can be crucial when dealing with other types of integrals. However, in our case, the integral was straightforward enough thanks to the piecewise function. The function \(\frac{|x|}{x}\) was already decomposed, allowing for direct integration across the interval \([a,b]\) without necessitating integration by parts.

Integral Bounds

In calculus, integral bounds define the limits between which a function is integrated. They are key in determining the area under the curve of the function. Integral bounds are specifically denoted in the definite integral by the lower bound \(a\) and the upper bound \(b\).
This integral operation calculates the net area between the function and the x-axis from \(a\) to \(b\). In our example, since the integral is defined between \(a\) and \(b\) with \(a < 0 < b\), we broke the integral into two parts:

• From \(a\) to \(0\), where the function evaluates to -1.
• From \(0\) to \(b\), where the function evaluates to 1.

This method ensures that each part of the piecewise-defined domain is handled correctly, allowing for precise evaluation of the piecewise function's integral.