Review question

# Can we find the area enclosed by this quadratic and the $x$-axis? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R9138

## Solution

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 $\tfrac{1}{3}$.