To determine

To evaluate:

∫0πfxdx 

where

fx= sin⁡x  if  0≤ x< π/2cos⁡x  if   π/2 ≤ x≤π

Answer

∫0πfxdx=0 

Explanation

1) Concept:

The Fundamental Theorem of Calculus: Suppose f is continuous on [a, b], then

∫abfxdx=Fb-F(a), where F is antiderivative of  f, that is F'=f.

2) Calculation:

∫0πfxdx= ∫0π/2f(x)⁡ dx+∫π/2πf(x) dx

By substituting the values of fx

∫0π/2sin⁡x dx+∫π/2πcos⁡x dx

By using the antiderivative of each term,

-cosθπ/20+[sin⁡θ]ππ/2

Applying the Fundamental Theorem of Calculus,

-cos⁡π2--cos⁡0+{sin⁡π-sin⁡π2} 

After substituting values of cos⁡0,sin⁡π,sin⁡π/2 and cos⁡π2,

-0--1+0-1=1-1=0

Conclusion:

Therefore,

∫0πfxdx=0