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.

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.

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

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.