Question

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

Answer 1

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`

Answer 2

`1+sqrt2`

Explanation:

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

Answer 3

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

Explanation:

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

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`