Review question

# What's the area enclosed by $y=x(x-3)^2$ and a tangent? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R7386

## Solution

Show that the gradient of the curve $y=x(x-3)^2$ is zero at the point $P(1,4)$, and sketch the curve.

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*}
If you know the product and chain rules, you could find the derivative using those instead. \begin{align*} \frac{dy}{dx}&{}=(x-3)^2+2x(x-3) \\ &{}= (x-3)(x-3+2x) \\ &{}= 3(x-1)(x-3). \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 $P$ cuts the curve again at $Q$. Calculate the areabetween the chord $PQ$ and the curve.

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 $(x-1)^2(x-4)=0.$

We could have written this down immediately, since the factorisation must be $(x-1)^2(x-a)$ for some $a$, which will expand to $x^3+px^2+qx-a$ for some $p$ and $q$, and so $a$ must be $4$ to match the constant terms.

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 $12-\dfrac{21}{4}=\dfrac{27}{4}=6\dfrac{3}{4}$.