Computing the value of an integral from Apostol I

Press90
Messages
3
Reaction score
0

Homework Statement


The notation [x] denotes the greatest integer <= x


"integral sign" with a lower limit of -1 and a upper limit of 3 [x] dx

Homework Equations


I would like to know how to enter symbols for certain problems.

I am lost when Apostol says "the notation [x] denotes the greatest integer <= x", i partially understand what he means.


The Attempt at a Solution



n/a
 
Physics news on Phys.org
the notation [x] denotes the greatest integer <= x; means exactly what is said.
[x] denotes the greatest integer less than or equal to x.
[2.1] = 2 [2.9] =2

Because 2 is the largest integer less than 2.1 and 2.9.The function when plotted looks like a staircase. You can compute the integral simply by adding the areas underneath the step; accounting for negative and positive area, ofcourse.

The integral should be something like 2 if I am not mistaken.
 
Let me give you a concrete example to illustrate [x]

[3.4] means the greatest integer that is less than or equal to 3.4 which is 3

The equal to is used when x is an integer itself, so

[3] = 3
 
thanks for the explanation guys
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

Similar threads

Back
Top