#int_0^pi max(sin(x),cos(x)) dx# =?

3 Answers
Apr 12, 2017

Assuming I've read the expression correctly: #color(green)(0)#

Explanation:

#0^pi=0#
So
#color(white)("XXX")intcolor(white)("x")0^picolor(white)("x")max(sin(x),cos(x))color(white)("x")dx=int color(white)("x")0 color(white)("x")dx =0#

Apr 12, 2017

#1+sqrt2#

Explanation:

#int_0^pimax(cosx,sinx)dx=int_0^(pi/4)cosxdx + int_(pi/4)^pi sinx dx = 1+sqrt2#

Apr 12, 2017

I get #1+sqrt2#. Please see below.

Explanation:

Here are the graphs of sine (red) and cosine (blue).

enter image source here

On #[0,pi]#, we have

#max(sinx,cosx) = {(sinx,"if", 0 <= x <= pi/4),(cosx,"if",pi/4 < x <= pi):}#

Therefore,

#int_0^pi max(sinx,cosx) dx = int_0^(pi/4) sinx dx + int_(pi/4)^pi cosx dx#

# = [sqrt2/2]+[1+sqrt2/2] = 1+sqrt2#