integration

Question

We have to find the integration of ∫2016₀  [arctanx]dx the limits are from 0 to 2016 . 

[.] means greastest integer

Answer 1

The key to doing this problem is to realize that the integrand is NOT a CONTINUOUS function; it's a step function with the x values at which the y values jump up (by 1) occurring at strange times, so it's awkward to try to figure out how many rectangles this could be made into.  For early values of x, arctan will be zero so y values will be 0.

 

Suffice it to say that as x goes from 0 to 2016, arctan varies from 0 to pi/2 so the largest y value is int(3.141592654.../2)=int(1.57) = 1.

 

Arctan x = 1 means that x = 1.107148 rad (63.43^o) and then int(arctan 1.107148) = 1, therefore

for every x in [1.107148, 2016], the y value will equal 1.

 

So the total area would be 1 (height) times (2016 - 1.107148),  or about 2014.9.

 

I do not think that analytical integration techniques are any good for this problem, because of its STEP Function behavior (non-continuity). You can't ignore the "greatest integer function" applied to the integrand. All you can do is empirically find the area under all the steps.

 

 

Answer 2

Michael, Mark,

 

Use integration by parts where you have two functions u and v, take the derivative of uv, and then integrate. In this case you end up with the integral of udv and vdu. 

 

With inverse trig, you can set  integration by parts where u is arctan x and dv is 1 (v=x). It works for arccos and arccsc too! 

Answer 3

arctanx can written as   tan^-1(x) = cosx /sinx

 

The integral is    F(2016) - F(0)

 

where       F(x) = ∫cosx(sinx)^-1 dx

 

 

Use substitution.

 

Let          u = sinx

              du = cosxdx

 

 

∫u^-1du =

 

-u^-2 + C =

 

-sin^-2x + C =

 

[-1 / sin²x ] + C

 

 

Now you can plug in the bounds into the integral.