no idea how to solve it

[Fuzell Wood]

For a real number x, let [x] denote the largest integer not exceeding x. (For example, [9/4]=2)
Evaluate [MATH]\Large \displaystyle \int\limits_ 1^10\ x(x-[x])dx[/MATH]the limits are 10 and 1
i didn't managed to write 10 in the top but i hope it is clear
Thank you in advance
I'm not native speaker of english,so sorry for any mistakes and so

 

[pka]

For a real number x, let [x] denote the largest integer not exceeding x. (For example, [9/4]=2)
Evaluate [MATH]\Large \displaystyle \int\limits_ 1^10\ x(x-[x])dx[/MATH]the limits are 10 and 1

See the graph HERE.

Here is the actual calculation.

 

[HallsofIvy]

To get more than one digit or character in a superscript (or subscript) put them in braces. Writing "^{10}" rather than "^10" will give \(\displaystyle \int_0^{10} x(x- [x])dx\).

Now, to do that integral break it into 10 parts:
\(\displaystyle \int_0^1 x(x- [x])dx+ \int_1^2 x(x- [x])dx+ \int_2^3 x(x- [x])dx+ \cdot\cdot\cdot+ \int_8^9 x(x- [x])dx+ \int_9^{10}x(x- [x]dx\)

which is
\(\displaystyle \int_0^1 x(x- 0)dx+ \int_1^2 x(x- 1)dx+ \int_2^3 x(x- 2)dx+ \cdot\cdot\cdot+ \int_8^9 x(x- 8)dx+ \int_9^{10} x(x- 9)dx\)

 

[Fuzell Wood]

To get more than one digit or character in a superscript (or subscript) put them in braces. Writing "^{10}" rather than "^10" will give \(\displaystyle \int_0^{10} x(x- [x])dx\).

Now, to do that integral break it into 10 parts:
\(\displaystyle \int_0^1 x(x- [x])dx+ \int_1^2 x(x- [x])dx+ \int_2^3 x(x- [x])dx+ \cdot\cdot\cdot+ \int_8^9 x(x- [x])dx+ \int_9^{10}x(x- [x]dx\)

which is
\(\displaystyle \int_0^1 x(x- 0)dx+ \int_1^2 x(x- 1)dx+ \int_2^3 x(x- 2)dx+ \cdot\cdot\cdot+ \int_8^9 x(x- 8)dx+ \int_9^{10} x(x- 9)dx\)

Thank you so much! Can you tell me where i can find this formula and explanation? What should i search for?

 

[HallsofIvy]

Find what formula? The fact that \(\displaystyle \int_a^b f(x)dx= \int_a^c f(x)dx+ \int_c^b f(x)dx\) comes from the basic definition of the integral.

 

[Fuzell Wood]

Find what formula? The fact that \(\displaystyle \int_a^b f(x)dx= \int_a^c f(x)dx+ \int_c^b f(x)dx\) comes from the basic definition of the integral.

i mean i thought that there is a formula that is used when it comes to evaluations with square brackets
sorry i haven't been taught higher maths, i try to study it on my own so that's why i'm such a slowpoke?

 

[Dr.Peterson]

The idea is to break the problem into cases (x between 0 and 1, between 1 and 2, ...), in each of which [x] is a constant. Do you see how HallsofIvy did that?

 

[Fuzell Wood]

The idea is to break the problem into cases (x between 0 and 1, between 1 and 2, ...), in each of which [x] is a constant. Do you see how HallsofIvy did that?

yes,i see