Integration Problem?

Question

Take the integral from [0 to pie] of 1/2(cos x + | cos x | ) dx =

and yes those are absolute value indications around the 2nd cos x.
0 = lower bound pie = upper bound

Answer 1

Integrate over the interval [0, π].

∫[(1/2)(cos x + | cos x |)] dx
________

Break the interval into two pieces then integrate.

Over the interval [0, π/2] cos x is positive. So:
(1/2)(cos x + | cos x |) = (1/2)(cos x + cos x) = cos x

Over the interval [π/2, π] cos x is negative. So:
(1/2)(cos x + | cos x |) = (1/2)(cos x - cos x) = 0
_________

∫[(1/2)(cos x + | cos x |)] dx | [Evaluated from 0 to π]

= ∫[(1/2)(cos x + | cos x |)] dx | [Evaluated from 0 to π/2] + 0

= ∫(cos x) dx | [Evaluated from 0 to π/2]

= sin x | [Evaluated from 0 to π/2]

= sin(π/2) - sin(0) = 1 - 0 = 1

Answer 2

No, it's spelled "pi". "Pie" is a tasty dessert, especially pumpkin pie. Moving on:

[0, π]∫1/2 (cos x + |cos x|) dx

First, break this into two integrals -- one from 0 to π/2, and the other from π/2 to π:

[0, π/2]∫1/2 (cos x + |cos x|) dx + [π/2, π]∫1/2 (cos x + |cos x|) dx

Now, everywhere on the interval [0, π/2], cos x is nonnegative, so |cos x| = cos x. Conversely, everywhere on the interval [π/2, π], cos x is nonpositive, so |cos x| = -cos x. So we can replace |cos x| with cos x in the first integral, and |cos x| with -cos x in the second, to obtain:

[0, π/2]∫1/2 (cos x + cos x) dx + [π/2, π]∫1/2 (cos x - cos x) dx
[0, π/2]∫cos x dx + [π/2, π]∫0 dx
[0, π/2]∫cos x dx
sin x |[0, π/2]
sin (π/2) - sin 0
1 - 0
1

And we are done.