What's the area enclosed by $y=x(x-3)^2$ and a tangent?

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Our plan is to differentiate to find the gradient of the curve. We begin by expanding \(y=x(x-3)^2\), giving \[y=x(x^2-6x+9)=x^3-6x^2+9x\] so \[\begin{align*} \frac{dy}{dx}&{}=3x^2-12x+9\\ &{}=3(x^2-4x+3)\\ &{}=3(x-3)(x-1). \end{align*}\]

Thus \(\dfrac{dy}{dx}=0\) when \(x=1\) and when \(x=3\). At \(x=1\), \(y=4\); at \(x=3\), \(y=0\).

So the gradient of the curve is zero at \(P(1,4)\) and also at \((3,0)\).

To sketch the curve, we note that the roots occur at \(x=0\) and \(x=3\) (with the latter being a double root, and so the curve touches the \(x\)-axis here). The \(y\)-intercept is \(0\).

The equation is a cubic with \(x^3\) as the leading term. So as \(x \to\infty\), \(y\to\infty\), and as \(x\to-\infty\), \(y\to-\infty\).

So the stationary point at \(P(1,4)\) must be a maximum, and that at\((3,0)\) must be a minimum, giving

The tangent at \(P\) has equation \(y=4\). This intersects the cubic again when \[4 = x(x-3)^2.\] We can expand this and rearrange to get \[x^3-6x^2+9x-4=0.\] We already know that \(x=1\) is a double root of this equation (as the line \(y=4\) touches the curve at \(x=1\)), so \((x-1)^2\) will be a factor. We can factorise to give \[(x-1)(x^2-5x+4)=0\] so \[(x-1)^2(x-4)=0.\]

Thus the tangent at \(P\) intersects the curve again at \(Q(4,4)\). We can now calculate the required area, indicated by \(R\) on this sketch.

The area of \(R\) is equal to the area of the rectangle between \(PQ\) and the \(x\)-axis, which is \(3\times4=12\), minus the area under the curve between \(P\) and \(Q\), which is given by \[\begin{align*} \int_1^4 x^3-6x^2+9x\,dx &{}= \left[\frac{x^4}{4}-2x^3+\frac{9x^2}{2}\right]_1^4 \\ &{}= (64-128+72)-\left(\frac{1}{4}-2+\frac{9}{2}\right) \\ &{}= 8-\frac{11}{4} = \frac{21}{4}. \end{align*}\]

Thus the area of \(R\) is \(12-\dfrac{21}{4}=\dfrac{27}{4}=6\dfrac{3}{4}\).