The area of the region bounded by the curve y = max{|x|, x|x - 2|}, then x-axis and the lines x = -2

Q: Why do we split the integral into four parts? Can't we just integrate the whole thing at once?

The 'max' function is not a single algebraic expression. It changes its identity whenever the curves cross. You must identify the 'upper envelope' in each interval because the function definition changes at the intersection points $x=0, 1, 3$.

Q: How do I handle the absolute value in $x|x-2|$ without getting confused?

Always define the function piecewise first. If $x \ge 2$, it is $x(x-2) = x^2 - 2x$. If $x < 2$, it is $x(2-x) = 2x - x^2$. This removes the ambiguity and allows you to treat the function as a simple polynomial in each sub-interval.

Q: Is there a geometric shortcut for these areas?

For the linear parts, yes, you can use the area of triangles. But for the parabolic parts, integration is the most robust method. Don't risk a geometric error when calculus is so reliable. Use the integration method to ensure precision.

Question

The area of the region bounded by the curve $y = \max\{|x|, x|x - 2|\}$, then x-axis and the lines $x = -2$ and $x = 4$ is equal to _______ .

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are tasked with finding the area bounded by the curve $y = \max\{|x|, x|x-2|\}$ between $x = -2$ and $x = 4$. Let us define our two functions: $f(x) = |x|$ and $g(x) = x|x-2|$.

The function $f(x)$ is the familiar V-shaped graph. For $x \ge 0$, it is $y = x$, and for $x < 0$, it is $y = -x$.

For $g(x) = x|x-2|$, we break it down based on the critical point $x = 2$:
*   When $x \ge 2$, $g(x) = x(x-2) = x^2 - 2x$.
*   When $x < 2$, $g(x) = x(2-x) = 2x - x^2$.

### Determining the Upper Envelope

The 'max' operator requires us to identify the upper envelope of these two curves. We find the intersection points by setting $|x| = x|x-2|$.

Solving this equation yields intersection points at $x = 0$, $x = 1$, and $x = 3$. These points partition our domain $[-2, 4]$ into four distinct intervals: $[-2, 0]$, $[0, 1]$, $[1, 3]$, and $[3, 4]$.

### Evaluating the Intervals

We determine which function is greater in each interval to set up our integrals:

*   In $[-2, 0]$, $f(x) = -x$ is above $g(x) = 2x - x^2$.
*   In $[0, 1]$, $g(x) = 2x - x^2$ is above $f(x) = x$.
*   In $[1, 3]$, $f(x) = x$ is above $g(x) = 2x - x^2$ (for $x \in [1, 2]$) and $g(x) = x^2 - 2x$ (for $x \in [2, 3]$).
*   In $[3, 4]$, $g(x) = x^2 - 2x$ is above $f(x) = x$.

### The Master Calculation

We now compute the area for each region:

$$Area_1 = \int_{-2}^{0} (-x) dx = \left[ -\frac{x^2}{2} \right]_{-2}^{0} = 0 - (-2) = 2$$

$$Area_2 = \int_{0}^{1} (2x - x^2) dx = \left[ x^2 - \frac{x^3}{3} \right]_{0}^{1} = 1 - \frac{1}{3} = \frac{2}{3}$$

$$Area_3 = \int_{1}^{3} x dx = \left[ \frac{x^2}{2} \right]_{1}^{3} = \frac{9}{2} - \frac{1}{2} = 4$$

$$Area_4 = \int_{3}^{4} (x^2 - 2x) dx = \left[ \frac{x^3}{3} - x^2 \right]_{3}^{4} = \left( \frac{64}{3} - 16 \right) - (9 - 9) = \frac{16}{3}$$

### Final Result

Summing these individual areas together, we obtain the total area:

$$Total Area = 2 + \frac{2}{3} + 4 + \frac{16}{3} = 6 + \frac{18}{3} = 6 + 6 = 12$$

The total area bounded by the curve is **12**.

Animated Solution for Mathematics - Definite Integration: The area of the region bounded by the curve $y = max {∣ x ∣, x ∣ x - 2∣}$ , then x-axis and the lines $x = - 2$ and $x = 4$ is equal to _______ .

Enter Numerical Value:

Visualized Solution

Analyze the Functions

Define $f (x) = ∣ x ∣$ Piecewise

Define $g (x) = x ∣ x - 2∣$ Piecewise

The Maximum Function

Compare in Interval $[- 2, 0]$

Compare in Interval $[0, 1]$

Compare in Interval $[1, 3]$

Compare in Interval $[3, 4]$

Integration Setup

Calculate Linear Areas $A_{1}$ and $A_{3}$

Calculate Parabolic Areas $A_{2}$ and $A_{4}$

Final Summation

The Sigma Insight: Area Bounded by Curves

Analyzing the Setup

Determining the Upper Envelope

Evaluating the Intervals

The Master Calculation

Final Result

Similar Questions

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Find all maxima and minima of the function $y = x (x - 1)^{2}, 0 \leq x \leq 2$ . Also determine the area bounded by the curve $y = x (x - 1)^{2}$ , the $y$ -axis and the line $y = 2$ .

Enter Numerical Value:

Visualized Solution

Analyze the Functions

Define .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)=∣x∣ Piecewise

Define .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g(x)=x∣x−2∣ Piecewise

The Maximum Function

Compare in Interval .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } [−2,0]

Compare in Interval .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } [0,1]

Compare in Interval .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } [1,3]

Compare in Interval .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } [3,4]

Integration Setup

Calculate Linear Areas .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A1​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A3​

Calculate Parabolic Areas .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A2​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A4​