Stranded on an Island

The graph of the function is composed of steps; a block of width 1 and height 4 from 0 to 1 and so on. Then the integral is simply the sum of the squares from 4 to 121 inclusive. Therefore the answer is $\boxed{505}$ .

The reason many people get 573 1/3 is because they integrated $\int _{x=0}^{10}{(x+2)^2 dx}$ not $\int _{ x=0 }^{ 10 }{ { \left\lfloor x+2 \right\rfloor }^{ 2 } dx }$ as wanted in the question.

The rule of thumb for integration involving any floor, ceiling, modulus functions is never to treat them as brackets and never to continue integrating as normal.

For these type of integrals it is best to draw the graph of the integrand, and then rewrite the integrals without using of the abovementioned types of functions before proceeding to integrate.

So the solution is as such: Draw the graph of $f(x)=\left\lfloor x+2 \right\rfloor$ from $x=0$ to $x=10$ and observe that $f(x)=(n+2)^2$ for $n<=x<n+1$ and $n$ is an integer. Hence notice that the integral is the area of 10 rectangles all of width 1 and height $2^2$ , $3^2$ , $4^2$ , ..., $10^2$ and $11^2$ respectively. Therefore the value of the integral $= 2^2+3^2+4^2+...+10^2+11^2=505$ .

Hope this would clear any conceptual misunderstanding that many appear to have.

I think there are already some fine approaches suggested. I played with the problem this way:

If $i \geq 0$ is some integer, we note that

$\int_{i}^{i+1} \! \lfloor x + 2 \rfloor ^2 \, \mathrm{d} x = \int_{i}^{i+1} \! (i + 2)^2 \, \mathrm{d} x$

So for some integer $n \geq 0$ , we get

$\int_{0}^{n} \! \lfloor i + 2 \rfloor ^2 \, \mathrm{d} x = \sum_{i=0}^{n-1} \int_{i}^{i+1} \! (i + 2)^2 \, \mathrm{d} x = \sum_{i=0}^{n-1} (i + 2)^2$

Recall the formula for pyramid numbers, and rewrite the sum

$\sum_{i=0}^{n-1} (i + 2)^2 = \frac{1}{6} n (2 n^2 + 9 n + 13)$

For the question at hand, we let $n = 10$ and get the result $505$ .

A natural extension of the above would be: "What would happen if $n$ were some real number?"

Am I missing something here? Expand then integrate gets (1/3)x^3+2x^2+4x which gives 573 1/3 when evaluated.

Here's a graph of the integrand.

Made using Sage

$\int_0^{10} {\lfloor {x+2}\rfloor}^2dx = {\sum\limits_{k=2}^{11} {k^2}}-1=\frac {11(11+1)(2\times {11} + 1)} 6 - 1 = \boxed {505}$

(1) (floor(0+2)^2) = 4 (1) (floor(1+2)^2) = 9 (1) (floor(2+2)^2) = 16 (1) (floor(3+2)^2) = 25 (1) (floor(4+2)^2) = 36 (1) (floor(5+2)^2) = 49 (1) (floor(6+2)^2) = 64 (1) (floor(7+2)^2) = 81 (1) (floor(8+2)^2) = 100 (1) (floor(9+2)^2) = 121 Sum of area of rectangles = 50