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