Signed area and absolute values

A definite integral is not an area but a signed area. We confirm it by integrating $y = x^3$ from $-1$ to $1$ .

\int_{-1}^{1} x^3 \, dx = \left[ \frac{x^4}{4} \right]_{-1}^{1} = \frac{1}{4} - \frac{1}{4} = 0

The answer is $0$ . Yet the graph plainly encloses some area. On $0 \leq x \leq 1$ the curve is above the $x$ -axis and on $-1 \leq x \leq 0$ below it, and the lower part was counted negative and cancelled the other.

When an area is what is wanted

Take the absolute value before integrating.

\int_{-1}^{1} |x^3| \, dx = 2\int_0^1 x^3 \, dx = 2 \times \frac{1}{4} = \frac{1}{2}

Since $|x|^3$ is even, computing the right half and doubling is enough.

Interval	$\int x^3 dx$	Area
$-1 \leq x \leq 0$	$-\dfrac{1}{4}$	$\dfrac{1}{4}$
$0 \leq x \leq 1$	$\dfrac{1}{4}$	$\dfrac{1}{4}$
Total	$0$	$\dfrac{1}{2}$

Using symmetry

Type of function	Integral over a symmetric interval
Odd	$0$
Even	twice one side

$x^3$ is odd and $|x|^3$ is even, so the two computations above are the typical cases. Knowing the symmetry shortens the work.

Why the sign is a virtue

Carrying a sign is not a defect. Integrating a velocity adds what was covered going forward and subtracts what was covered coming back, leaving the net displacement. Use the absolute value when an area is wanted, and integrate as it stands when a net change is wanted.

Seen from the antiderivative

The antiderivative $\dfrac{x^4}{4}$ of $x^3$ is even, so it takes the same value at $x = -1$ and $x = 1$ ; the difference being $0$ is only to be expected. The antiderivative of $|x|^3$ , by contrast, is $\dfrac{x^4}{4}$ for $x \geq 0$ and $-\dfrac{x^4}{4}$ for $x < 0$ : the formula changes with the interval.

To find an area, first look for the $x$ with $f = 0$ and split the interval there. For this function the dividing point is $x = 0$ .

The curve symmetric about the origin is $y = x^3$ , the curve symmetric about the $y$ -axis is $y = |x|^3$ , and the large dots are $(-1, -1)$ , $(0, 0)$ and $(1, 1)$ .