Chapter 8 · Application of Integrals

Miscellaneous · Q4 (MCQ): Area between $y = x^3$ ⁠, the $x$ -axis, $x = -2$ and $x = 1$

Miscellaneous ExerciseA curve below the axis: split where it crosses · Miscellaneous Exercise

The question

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

4. Area bounded by the curve $y = x^3$ ⁠, the $x$ -axis and the ordinates $x = -2$ and $x = 1$ is

(A) $-9$ (B) $\dfrac{-15}{4}$ (C) $\dfrac{15}{4}$ (D) $\dfrac{17}{4}$

Exam answer

(figure) The curve $y = x^3$ meets the $x$ -axis at $O(0, 0)$ ⁠, the line $x = -2$ at $(-2, -8)$ and the line $x = 1$ at $(1, 1)$ ⁠.

$y = x^3 < 0$ for $-2 \le x < 0$ and $y \ge 0$ for $0 \le x \le 1$ ⁠.

\begin{aligned} A &= \left|\int_{-2}^{0}x^3\,dx\right| + \int_0^1 x^3\,dx \\ &= \left|\left[\frac{x^4}{4}\right]_{-2}^{0}\right| + \left[\frac{x^4}{4}\right]_0^1 \\ &= |0 - 4| + \Big(\frac14 - 0\Big) \end{aligned}

Answer: (D) $\dfrac{17}{4}$ sq. units

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

The one idea this problem is built on

The cubic $y = x^3$ crosses the $x$ -axis at $x = 0$ ⁠, which lies between the ordinates $x = -2$ and $x = 1$ ⁠. To the left of 0 the curve is below the axis and its integral is negative ( $-4$ ⁠); to the right it is above ( $\frac14$ ⁠). The area is the sum of the sizes, $4 + \frac14 = \frac{17}{4}$ ⁠, option (D). The other options are what you get by integrating in one go ( $-\frac{15}{4}$ ⁠), by taking the size only at the end ( $\frac{15}{4}$ ⁠), or a negative number, which no area can be.

Given, to find, and the plan

Given: The curve $y = x^3$ ⁠, the $x$ -axis, and the ordinates (vertical lines) $x = -2$ and $x = 1$ ⁠.
To find: The area of the region they bound, and which option it is.
Plan: (1) Find where the curve crosses the axis and its sign on each side; sketch. (2) Split at the crossing and put bars on the part below the axis. (3) Evaluate both parts and add. (4) Examine every option.

Basics you need

1Area below the axis

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

2Sign of a cube

$x^3$ has the same sign as $x$ ⁠: a negative number cubed is negative, for example $(-2)^3 = -8$ ⁠.

3Power rule and the definite integral

$\int x^3\,dx = \frac{x^4}{4} + C$ (Chapter 7, Section 7.2), and $\int_a^b f = F(b) - F(a)$ ⁠. Basics: the definite integral

4Split at the crossing

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

Solution

Where is the curve below the axis?

$x^3 = 0$ only at $x = 0$ ⁠, which is inside $[-2, 1]$ ⁠. Since $x^3$ has the sign of $x$ ⁠:
- for $-2 \le x < 0$ ⁠, $y = x^3 < 0$ ⁠: the curve is below the axis, down to $(-2, -8)$ ⁠;
- for $0 < x \le 1$ ⁠, $y = x^3 > 0$ ⁠: the curve is above the axis, up to $(1, 1)$ ⁠.
$y = x^3$ between the ordinates $x = -2$ and $x = 1$ ⁠. The rose region (below the axis) has signed integral $-4$ and area 4; the small blue region (above) has area $\frac14$ ⁠. The $y$ -axis is squeezed (one unit up is less than a quarter of one unit across) so that $(-2, -8)$ fits.

The crossing $x = 0$ splits the region into a large piece under the axis and a small piece over it. A single integral would let the small piece cancel part of the large one.

Lesson: a curve that dips below the axis
Write the area with bars on the lower part
$A = \left|\int_{-2}^{0}x^3\,dx\right| + \int_0^1 x^3\,dx .$
Split at the crossing (P2), then take the size of the part below the axis, as in the book's rule $A = |A_1| + A_2$ ⁠.

Basics: area below the axis
Evaluate the two parts
$\begin{aligned} \int_{-2}^{0}x^3\,dx &= \left[\frac{x^4}{4}\right]_{-2}^{0} = 0 - \frac{(-2)^4}{4} \\ &= 0 - \frac{16}{4} = -4 , \end{aligned}$ $\int_0^1 x^3\,dx = \left[\frac{x^4}{4}\right]_0^1 = \frac14 - 0 = \frac14 .$
The power rule with $n = 3$ gives $\frac{x^4}{4}$ (its derivative is $x^3$ ⁠). Note $(-2)^4 = 16$ is positive (an even power), so the bracket on the left is $0 - 4$ ⁠. The negative result $-4$ confirms that this part is below the axis.

Basics: the definite integral is F(b) − F(a)
Add the sizes
$A = \lvert -4\rvert + \frac14 = 4 + \frac14 = \frac{17}{4} .$
The area is $\frac{17}{4}$ sq. units. This is option (D).

Each part now counts as a positive area, so nothing cancels.

$4 = \frac{16}{4}$ ⁠, so $4 + \frac14 = \frac{17}{4}$ ⁠.
Examine every option
- (A) $-9$ : negative, and an area is never negative. It is not even the signed integral, which is $-\frac{15}{4}$ ⁠. Wrong.
- (B) $\frac{-15}{4}$ : this is the single integral $\int_{-2}^{1}x^3\,dx = \frac14 - 4 = -\frac{15}{4}$ ⁠, the net signed area, where the part above the axis cancels part of the part below. Negative, so not an area. Wrong.
- (C) $\frac{15}{4}$ : this is $\left|\int_{-2}^{1}x^3\,dx\right|$ ⁠, the bars put round the whole integral. The cancellation has already happened: $4 - \frac14$ instead of $4 + \frac14$ ⁠. Wrong.
- (D) $\frac{17}{4}$ : the sum of the sizes of the two parts, as found in Step 4 ✓. Correct.
Options (B) and (C) are the two classic slips with a curve that crosses the axis. Only splitting at the crossing first, then taking sizes, gives the area.

See why the right option is right

Result

The area bounded by $y = x^3$ ⁠, the $x$ -axis and the ordinates $x = -2$ and $x = 1$ is

\begin{aligned} &\left|\int_{-2}^{0}x^3\,dx\right| + \int_0^1 x^3\,dx \\ &= 4 + \frac14 = \frac{17}{4}\ \text{sq. units}. \end{aligned}

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

Check it with numbers

A computer's numerical integration of $|x^3|$ over $[-2, 1]$ ⁠, with a break at 0, gives 4.2500000000, and $\frac{17}{4} = 4.25$ ✓. Of $x^3$ itself it gives $-4.0000000000$ over $[-2, 0]$ ⁠, 0.2500000000 over $[0, 1]$ and $-3.7500000000$ over $[-2, 1]$ ⁠, which is option (B)'s $-\frac{15}{4}$ ⁠.

Common mistakes

Integrating from $-2$ to 1 in one go and getting $-\frac{15}{4}$ ⁠, option (B).
Fixing the sign at the end: $\left|-\frac{15}{4}\right| = \frac{15}{4}$ ⁠, option (C). The bars must go on the part below the axis, before adding.
$(-2)^4 = -16$ : an even power of a negative number is positive.

What to take away

When the interval contains a point where $y = 0$ and the sign changes, split there and take sizes.
In MCQs on areas, a negative option and "the plain integral" are traps. The same pattern appears in Miscellaneous Q3 and Miscellaneous Q5.

Every fact used, and where it comes from

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

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Miscellaneous · Q4 (MCQ): Area between y=x3y = x^3y=x3⁠, the xxx-axis, x=−2x = -2x=−2 and x=1x = 1x=1

Basics you need

Solution

Where is the curve 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 · Q4 (MCQ): Area between $y = x^3$ ⁠, the $x$ -axis, $x = -2$ and $x = 1$