Let f(x) = max { |x+1|, |x+2|, ..., |x+5| }. Then ∫[-6 to 0] f(x) dx is equal to ______.

Q: Why can we just ignore $|x+2|, |x+3|, |x+4|$? Isn't that cheating?

It's not cheating; it's optimization! Think of a tent. The poles at the ends hold up the fabric. The intermediate poles are shorter and don't touch the canvas. Mathematically, for any $x$, the maximum of a set of functions is determined by the 'outer' functions. Since $|x+1|$ and $|x+5|$ are the furthest apart, they define the upper boundary.

Q: How do I solve $|x+1| = |x+5|$ without getting confused by the modulus?

Great question. Remember that $|a| = |b|$ implies $a = b$ or $a = -b$. Here, $x+1 = x+5$ gives $1=5$ (impossible), so we must use $x+1 = -(x+5)$. This leads to $2x = -6$, or $x = -3$. Always look for the case where the slopes have opposite signs!

Q: Why do we split the integral at $x = -3$?

Because the definition of $f(x)$ changes there! Before $x = -3$, the function follows one line; after $x = -3$, it follows another. You cannot integrate a piecewise function across its 'kink' without splitting it. Think of it as changing lanes on a highway—you have to adjust your steering!

Question

Let $f(x) = \max \{ |x+1|, |x+2|, \dots, |x+5| \}$. Then $\int_{-6}^{0} f(x) dx$ is equal to ______.

Sigma Percentile · Accepted Answer

### The Chaotic Forest of Lines

Imagine you are standing in a forest, but instead of trees, you are surrounded by five V-shaped graphs: $|x+1|$, $|x+2|$, $|x+3|$, $|x+4|$, and $|x+5|$.

At first glance, it looks like a chaotic, tangled mess. Your task is to find the area under the 'envelope' of these functions—the maximum value at any given point $x$—from $x = -6$ to $x = 0$.

This is the essence of our problem. We are looking for the integral:

$$\int_{-6}^{0} f(x) dx$$

where $f(x) = \max \{ |x+1|, |x+2|, |x+3|, |x+4|, |x+5| \}$.

### The Envelope Strategy

Do not be intimidated by the number of functions. In any set of absolute value functions, the maximum value is always dictated by the 'outermost' poles.

Think of a tent; the fabric is held up by the tallest poles at the ends. The intermediate poles, like $|x+2|$, $|x+3|$, and $|x+4|$, are shorter and simply sit beneath the fabric.

Therefore, the maximum function simplifies beautifully to just the maximum of the two extreme functions:

$$f(x) = \max \{ |x+1|, |x+5| \}$$

This is our 'orange envelope'—the boundary that defines our area.

### The Critical Junction

Now, we need to know where the transition happens. Where does the graph of $|x+1|$ stop being the maximum and hand over the baton to the graph of $|x+5|$?

We find this by setting them equal: $|x+1| = |x+5|$. To solve this, remember that for $|a| = |b|$, we must have $a = b$ or $a = -b$.

The equation $x+1 = x+5$ leads to $1 = 5$, which is impossible. So, we look at the other case: $x+1 = -(x+5)$.

Solving this gives $2x = -6$, or $x = -3$. This is our critical junction, the point where the function changes its behavior.

### Defining the Piecewise Function

With our boundary at $x = -3$, we can now define $f(x)$ piecewise.

For the interval $x \in [-6, -3]$, the function $|x+1|$ is the higher one. Since $x+1$ is negative in this range, $|x+1| = -(x+1) = -x-1$.

For the interval $x \in [-3, 0]$, the function $|x+5|$ takes over. Since $x+5$ is positive here, $|x+5| = x+5$.

Thus, our function is defined as:

$$f(x) = \begin{cases} -(x+1) & x \in [-6, -3] \ x+5 & x \in [-3, 0] \end{cases}$$

### The Integration

Now, we calculate the area. We split the integral into two parts:

$$I_1 = \int_{-6}^{-3} (-x-1) dx \quad 	ext{and} \quad I_2 = \int_{-3}^{0} (x+5) dx$$

For $I_1$, the anti-derivative of $-x-1$ is $-\frac{x^2}{2} - x$. Evaluating this from $-6$ to $-3$:

$$\left( -\frac{(-3)^2}{2} - (-3) ight) - \left( -\frac{(-6)^2}{2} - (-6) ight) = (-4.5 + 3) - (-18 + 6) = -1.5 - (-12) = 10.5$$

For $I_2$, the anti-derivative of $x+5$ is $\frac{x^2}{2} + 5x$. Evaluating this from $-3$ to $0$:

$$(0) - \left( \frac{(-3)^2}{2} + 5(-3) ight) = 0 - (4.5 - 15) = 10.5$$

Adding these together, $10.5 + 10.5 = 21$. We have conquered the forest! The final area is **21**.

Animated Solution for Mathematics - Definite Integration: Let $f (x) = max {∣ x + 1∣, ∣ x + 2∣, \dots, ∣ x + 5∣}$ . Then $\int_{- 6}^{0} f (x) d x$ is equal to ______.

Enter Numerical Value:

Visualized Solution

Understanding $f (x)$

Simplifying the Maximum

Finding the Intersection

Solving for $x$

Defining $f (x)$ Piecewise

Setting up the Integral

Integrating the First Part

Evaluating the First Area

Integrating the Second Part

Evaluating the Second Area

Final Summation

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

The Chaotic Forest of Lines

The Envelope Strategy

The Critical Junction

Defining the Piecewise Function

The Integration

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