Area Under the Curve
Area with Piecewise and Floor Functions
nta_pyq_2023_apr
Grade 12
Question:
Let the area enclosed by the lines $x+y=2$, $y=0$, $x=0$ and the curve $f(x)=\min\!\left\{x^2+\dfrac{3}{4},\ 1+[x]\right\}$ where $[x]$ denotes the greatest integer $\leq x$, be $A$. Then the value of $12A$ is
Step-by-Step Solution
Key Concept: On $[0,1)$: $1+[x]=1$ and $x^2+\frac{3}{4}\geq 1$ for $x\geq\frac{1}{2}$, so $f(x)=x^2+\frac{3}{4}$ for $x\in[0,\frac{1}{2})$ and $f(x)=1$ for $x\in[\frac{1}{2},1)$. On $[1,2]$: $f=2$, but the triangle $x+y=2$ limits integration.
$A=\int_0^{1/2}\!\left(x^2+\frac{3}{4}\right)dx+\int_{1/2}^1 1\,dx+\frac{1}{2}=\frac{5}{12}+1=\frac{17}{12}$. $12A=17$.
Correct Answer: 17