Let the area enclosed by the lines x+y=2, y=0, x=0, and the curve f(x)=(x^2+34, 1+[x]) (wh…

Question

Let the area enclosed by the lines $x+y=2$, $y=0$, $x=0$, and the curve $f(x)=\min\left(x^2+\dfrac34,\ 1+[x]\right)$ (where $[x]$ denotes the greatest integer $\le x$) be $A$. Then the value of $12A$ is \_\_\_\_\_\_\_.

Answer: 17

Accepted Answer

For $x\in[0,1)$: $1+[x]=1$; $x^2+\tfrac34<1 \iff x<\tfrac12$. So $f(x)=x^2+\tfrac34$ on $[0,\tfrac12)$ and $f(x)=1$ on $[\tfrac12,1)$ --- in both cases $f(x)<2-x$ there, so $f$ is the lower envelope.

For $x\in[1,2)$: $1+[x]=2$, while $x^2+\tfrac34\ge1.75>2-x$; here the line $y=2-x$ is the lower envelope.

$$A=\int_0^{1/2}\left(x^2+\frac34\right)dx+\int_{1/2}^1 1\,dx+\int_1^2(2-x)\,dx=\frac5{12}+\frac12+\frac12=\frac{17}{12}.$$

$12A=17.$

Solutions · 1