The curve \(y=ax^2+bx+c\) passes through the points \((1,0)\) and \((2,0)\) and its gradient at the point \((2,0)\) is \(2\). Find the numerical value of the area included between the curve and the axis of \(x\).

Since the curve passes through \((1,0)\) and \((2,0)\) we have \[\begin{align*} ax^2+bx+c &= a(x-1)(x-2) \\ &= ax^2 -3ax +2a \end{align*}\] Equating coefficients we find that \[\begin{align} b&=-3a \label{eq:b} \\ \text{and }c&=2a \label{eq:c} \end{align}\]

We also know the gradient of the curve is \(2\) at \(x=2\). Differentiating, \[ \frac{dy}{dx}=2ax+b, \] and so we have that \[ 4a+b=2. \] Substituting \(\eqref{eq:b}\) and \(\eqref{eq:c}\), we have \[ a=2,\quad b=-6,\quad c=4. \] So the curve has the equation \(y=2x^2-6x+4\).

The area between the curve and the axis is found by integrating between the two roots at \(x=1\) and \(x=2\). Since the curve is below the axis this will give us a negative result whose numerical value is the area. \[ \int_1^2 2x^2-6x+4 \, dx = \left[\frac{2x^3}{3}-3x^2+4x\right]^2_1=\frac{16}{3}-12+8-\frac{2}{3}+3-4=\frac{14}{3}-5=-\frac{1}{3} \] Thus the required area is \(\tfrac{1}{3}\).