Can we explain why this integral is zero?

Can we explain why this integral is zero?

Question

The curve $y = 4(x+1)(x-1)(x-3)$ meets the $x$-axis at three points. Verify, by finding the numerical value of each, that the two areas formed between the curve and the $x$-axis are of equal size.

Explain why the value of $\displaystyle \int_{-1}^3 y \:dx$ is zero.

Solution

The three intercepts with the $x$-axis are at $x = -1$, $x = 1$, and $x = 3$.

The curve has a positive $x^3$ coefficient, so we know the rough shape of the curve. The region enclosed between the curve and the $x$-axis between $x = -1$ and $x = 1$ is above the $x$-axis, while the region enclosed between $x = 1$ and $x = 3$ is below the $x$-axis.

We can expand the expression that defines

$y$

:

$$\begin{align*} y = 4(x+1)(x-1)(x-3) &= 4(x^2 - 1)(x-3) \\ &= 4(x^3 - 3x^2 - x + 3). \end{align*}$$

The first area is

$$\begin{equation*} \int_{-1}^1 y \:dx = 4 \int_{-1}^1 x^3 - 3x^2 - x + 3 \:dx = \left[ x^4 - 4x^3 - 2x^2 + 12x \right]_{-1}^{1} = (7)-(-9) = 16. \end{equation*}$$

We can find the second area by evaluating

$$\begin{align*} \left[ x^4 - 4x^3 - 2x^2 + 12x \right]_{1}^{3} = -16. \end{align*}$$

This is negative since the curve lies beneath the $x$-axis here. Hence the value of the second area is $16$, and the magnitudes of the two areas are equal, as required.

We can split

$$\begin{equation*} \int_{-1}^3 y \:dx = \int_{-1}^1 y \:dx + \int_1^3 y \:dx. \end{equation*}$$

We have seen that the two integrals are equal in magnitude, but opposite in sign (as the second region lies below the $x$-axis), and so the above quantity vanishes.