Calculus BC problem So I as attempting to solve this problem and just got stuck overall. Since this is just practice an explanation is much more valuable than the answer itself. Thanks!
Here is the problem


$$\int_0^\pi \sin(x)dx =- cos(x)[_0^\pi =2 $$
So the green part must have an area of 1 
Split the green part into ...
a right triangle with base $b=k$ and height $ h=sin(k)$ 
and
the area under the curve between $x=k$ and $x = \pi $
$$ \frac 12 k \sin(k) + \int_k^\pi \sin(x) dx = 1  $$
$$ \frac 12 k \sin(k) + 1-\cos(k) = 1  $$
