Area supplies another coefficient equation · source reference

Printed question

The diagram below shows a portion of a quadratic curve $y=f(x)$. The area of the shaded region bounded by the axes, the curve $y=f(x)$ and the line $x=1$ is $\frac{19}{6}$ units$^2$. Given also that the curve passes through the points $(1,\frac{13}{2})$ and $(-2,\frac12)$, determine the equation of the curve. [5]

[2011/MJC/Prelim/I/6b]

Printed working

Let $f(x)=ax^2+bx+c$.

Substitute $(1,\frac{13}{2})$ and $(-2,\frac12)$ into $y=f(x)$ to obtain:

$$a+b+c=\frac{13}{2}\qquad(1),$$ $$4a-2b+c=\frac12\qquad(2).$$ $$\int_0^1(ax^2+bx+c)\,dx=\frac{19}{6}.$$ $$\left[\frac{ax^3}{3}+\frac{bx^2}{2}+cx\right]_0^1=\frac{19}{6}.$$ $$\frac a3+\frac b2+c=\frac{19}{6}\qquad(3).$$

Using G.C. to solve equations (1), (2) and (3), $a=2,b=4,c=\frac12$.

Therefore equation of the curve is $y=2x^2+4x+\frac12$.

Teacher notes

Area of shaded region $=\int_0^1f(x)\,dx$.

Printed answer key

$y=2x^2+4x+\frac12$.

Source provenance

2025 TJC H2 Math (9758) JC2 MSM: EQUATIONS. Question sheet printed pages 15–19; solution sheet printed pages 1–12. Original exam attribution is retained above. Currency dollar signs are spelled out outside maths to avoid confusing currency with TeX delimiters.