Let f(x) = [x-1],[x-2],,[x-10], where [t] denotes the greatest integer t.

Question

Let $f(x) = \min\{[x-1],[x-2],\ldots,[x-10]\}$, where $[t]$ denotes the greatest integer $\le t$. Then $\displaystyle\int_0^{10}f(x)\,dx+\int_0^{10}\big(f(x)\big)^2dx+\int_0^{10}|f(x)|\,dx$ is equal to \_\_\_\_\_\_\_\_.

Answer: 385

Accepted Answer

Since $[x-k]$ is decreasing in $k$ for fixed $x$, the minimum is $f(x)=[x-10]$. On $[0,10]$, $f(x)\le0$, so $|f(x)|=-f(x)$, and

$$\int_0^{10}f(x)\,dx+\int_0^{10}|f(x)|\,dx = \int_0^{10}f(x)\,dx-\int_0^{10}f(x)\,dx=0$$

So the total reduces to $\displaystyle\int_0^{10}\big(f(x)\big)^2dx = \displaystyle\int_0^{10}[x-10]^2dx$, and breaking into unit intervals ($[x-10]=-k$ for $x\in[10-k,11-k)$, $k=1,\ldots,10$):

$$= 10^2+9^2+8^2+\cdots+1^2 = \frac{10\cdot11\cdot21}6 = 385$$

Solutions · 1