Chapter 8 · Application of Integrals

Miscellaneous · Q5 (MCQ): Area between $y = x|x|$ ⁠, the $x$ -axis, $x = -1$ and $x = 1$

Miscellaneous ExerciseA modulus inside the curve: split at the corner · Miscellaneous Exercise

The question

Choose the correct answer in the following Exercises from 4 to 5.

5. The area bounded by the curve $y = x\,|x|$ ⁠, $x$ -axis and the ordinates $x = -1$ and $x = 1$ is given by

(A) $0$ (B) $\dfrac{1}{3}$ (C) $\dfrac{2}{3}$ (D) $\dfrac{4}{3}$

[Hint : $y = x^2$ if $x > 0$ and $y = -x^2$ if $x < 0$ ].

Exam answer

(figure) The curve $y = x|x|$ meets the $x$ -axis at $O(0, 0)$ ⁠, the line $x = -1$ at $(-1, -1)$ and the line $x = 1$ at $(1, 1)$ ⁠.

$y = x|x| = -x^2$ for $x < 0$ and $y = x^2$ for $x \ge 0$ ⁠.

\begin{aligned} A &= \left|\int_{-1}^{0}(-x^2)\,dx\right| + \int_0^1 x^2\,dx \\ &= \left|\left[-\frac{x^3}{3}\right]_{-1}^{0}\right| + \left[\frac{x^3}{3}\right]_0^1 \\ &= \left|0 - \frac13\right| + \Big(\frac13 - 0\Big) \end{aligned}

Answer: (C) $\dfrac{2}{3}$ sq. units

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

Remove the bars: for $x \ge 0$ ⁠, $x|x| = x^2$ (above the axis); for $x < 0$ ⁠, $x|x| = -x^2$ (below the axis). So the region is two pieces that meet at the origin, each of area $\frac13$ ⁠. Splitting at $x = 0$ and taking sizes gives $\frac13 + \frac13 = \frac23$ ⁠, option (C). A single integral gives 0, because the lower piece cancels the upper one exactly.

Given, to find, and the plan

Given: The curve $y = x|x|$ ⁠, the $x$ -axis, and the ordinates $x = -1$ and $x = 1$ ⁠; the hint that $y = x^2$ for $x > 0$ and $y = -x^2$ for $x < 0$ ⁠.
To find: The area of the region they bound, and which option it is.
Plan: (1) Write $y$ without bars on each side of 0 (the hint). (2) Sketch and find where it is below the axis. (3) Split at 0, with bars on the lower part. (4) Evaluate and add. (5) Examine every option.

Basics you need

1Absolute value

$|x| = x$ for $x \ge 0$ and $|x| = -x$ for $x < 0$ ⁠.

2Area below the axis

Where $y < 0$ ⁠, the integral is negative and the area is its absolute value; $A = |A_1| + A_2$ (Section 8.2, Remark). Basics: area below the axis

3Power rule and the definite integral

$\int x^2\,dx = \frac{x^3}{3} + C$ (Chapter 7, Section 7.2), and $\int_a^b f = F(b) - F(a)$ ⁠. Basics: the definite integral

4Split at the corner

$\int_a^b = \int_a^c + \int_c^b$ (property P2, Chapter 7, Section 7.10), with $c$ where the formula changes. Lesson: below the axis, take the size

Solution

Write $y = x|x|$ without the bars
$x|x| = \begin{cases} x(-x) = -x^2, & x < 0,\\ x \cdot x = x^2, & x \ge 0.\end{cases}$
Both formulas give 0 at $x = 0$ ⁠. This is the book's hint.

With the bars gone, each side is a plain power of $x$ that the power rule can integrate.

Test: at $x = -\frac12$ ⁠, $x|x| = -\frac12 \cdot \frac12 = -\frac14$ and $-x^2 = -\frac14$ ✓.
Sketch: which part is below the axis?

For $-1 \le x < 0$ ⁠, $y = -x^2 < 0$ ⁠: below the axis, down to $(-1, -1)$ ⁠. For $0 < x \le 1$ ⁠, $y = x^2 > 0$ ⁠: above the axis, up to $(1, 1)$ ⁠. The curve crosses the axis only at the origin.

$y = x|x|$ drawn with equal scales: the right half of $y = x^2$ (blue) and the left half of $y = -x^2$ (rose). Between $x = -1$ and $x = 1$ the region has a blue piece above the axis and a rose piece below it, each of area $\frac13$ ⁠. Turning the picture half a turn about the origin swaps the two pieces.

The half turn shows the two pieces are congruent, because $y = x|x|$ is odd: $(-x)\lvert -x\rvert = -x|x|$ ⁠. Their signed integrals are equal and opposite, which is why a single integral gives 0.

Lesson: a curve that dips below the axis
Write the area with bars on the lower part
$A = \left|\int_{-1}^{0}(-x^2)\,dx\right| + \int_0^1 x^2\,dx .$
Split at the corner $x = 0$ (P2), use each side's formula, and take the size of the part below the axis.

Basics: area below the axis
Evaluate the two parts
$\begin{aligned} \int_{-1}^{0}(-x^2)\,dx &= \left[-\frac{x^3}{3}\right]_{-1}^{0} \\ &= 0 - \left(-\frac{(-1)^3}{3}\right) \\ &= 0 - \frac13 = -\frac13 , \end{aligned}$ $\int_0^1 x^2\,dx = \left[\frac{x^3}{3}\right]_0^1 = \frac13 - 0 = \frac13 .$
An antiderivative of $-x^2$ is $-\frac{x^3}{3}$ ⁠. At $x = -1$ ⁠: $(-1)^3 = -1$ ⁠, so $-\frac{(-1)^3}{3} = \frac13$ ⁠; subtracting it from the value 0 at $x = 0$ gives $-\frac13$ ⁠. The minus sign confirms the part is below the axis.

Basics: the definite integral is F(b) − F(a)
Add the sizes
$A = \left|-\frac13\right| + \frac13 = \frac13 + \frac13 = \frac23 .$
The area is $\frac23$ sq. units. This is option (C).

Each piece counts as a positive area, so they add instead of cancelling.

Basics: take the size
Examine every option
- (A) $0$ : this is the single integral $\int_{-1}^{1}x|x|\,dx = -\frac13 + \frac13 = 0$ (zero for any odd function, by property P7 (ii) of Chapter 7). It is the net signed area, not the area. Wrong.
- (B) $\frac13$ : the area of only one of the two pieces. Wrong.
- (C) $\frac23$ : the sum of the two sizes, as found in Step 5 ✓. Correct.
- (D) $\frac43$ : twice the correct area. It is, for example, the area between the two full parabolas $y = x^2$ and $y = -x^2$ from $-1$ to 1, $\int_{-1}^{1}2x^2\,dx = \frac43$ ⁠, a different region. Wrong.
Option (A) is the trap: a region can have a positive area while its signed integral is 0.

See why the right option is right

Result

The area bounded by $y = x|x|$ ⁠, the $x$ -axis and the ordinates $x = -1$ and $x = 1$ is

\begin{aligned} &\left|\int_{-1}^{0}(-x^2)\,dx\right| + \int_0^1 x^2\,dx \\ &= \frac13 + \frac13 = \frac23\ \text{sq. units}. \end{aligned}

The correct option is (C). (Textbook answer: C.)

Check it with numbers

A computer's numerical integration of $\big|x|x|\big|$ over $[-1, 1]$ ⁠, with a break at 0, gives 0.6666666667, and $\frac23 \approx 0.6666666667$ ✓. Of $x|x|$ itself it gives $-0.3333333333$ over $[-1, 0]$ ⁠, 0.3333333333 over $[0, 1]$ and 0.0000000000 over $[-1, 1]$ ⁠, which is option (A).

A second way to see the answer: the size of the height is $\big|x|x|\big| = |x|^2 = x^2$ for every $x$ ⁠, so the area is also $\int_{-1}^{1}x^2\,dx = \frac13 - \left(-\frac13\right) = \frac23$ ✓.

Common mistakes

Integrating from $-1$ to 1 in one go and getting 0, option (A).
Putting the bars round the whole integral: $\left|\int_{-1}^{1}x|x|\,dx\right| = |0| = 0$ ⁠. The bars belong on the part below the axis, after splitting.
Sign slip at $x = -1$ : $-\frac{(-1)^3}{3} = +\frac13$ ⁠, so the left integral is $0 - \frac13$ ⁠.

What to take away

A modulus inside the curve is removed by splitting where its inside is zero, as in Miscellaneous Q2.
For an odd function on $[-a, a]$ ⁠, the integral is 0 but the area is twice the area of one side.

Every fact used, and where it comes from

Fact	Used in	Source
$\|x\| = x$ for $x \ge 0$ ⁠, $-x$ for $x < 0$	Step 1	Definition of absolute value; the question's hint
Part below the axis: take the absolute value; $A = \|A_1\| + A_2$	Steps 3 and 5	Section 8.2, Remark (Figs 8.3 and 8.4) (Basics)
P2: $\int_a^b = \int_a^c + \int_c^b$	Step 3	Chapter 7, Section 7.10 (Lesson)
$\int x^n\,dx = \frac{x^{n+1}}{n+1} + C$	Step 4	Chapter 7, standard formula (i), Section 7.2
$\int_a^b f(x)\,dx = F(b) - F(a)$	Steps 4 and 6	Chapter 7, Theorem 2, Section 7.8.3 (Basics)
P7 (ii): $\int_{-a}^{a}f = 0$ for odd $f$	Step 6	Chapter 7, Section 7.10

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Miscellaneous · Q5 (MCQ): Area between y=x∣x∣y = x|x|y=x∣x∣⁠, the xxx-axis, x=−1x = -1x=−1 and x=1x = 1x=1

Basics you need

Solution

Write y=x∣x∣y = x|x|y=x∣x∣ without the bars

Sketch: which part is below the axis?

Write the area with bars on the lower part

Evaluate the two parts

Add the sizes

Examine every option

Every fact used, and where it comes from

Miscellaneous · Q5 (MCQ): Area between $y = x|x|$ ⁠, the $x$ -axis, $x = -1$ and $x = 1$

Write $y = x|x|$ without the bars