Asked by Scott
                Find the area of the region between y = x sin x and y = x for 0 ≤ x ≤ pi divided by 2
            
            
        Answers
                    Answered by
            MathMate
            
    You will need to integrate the function
f(x)=xsin(x)-x
=x(sin(x)-1)
between 0 and π/2.
Since we are looking for the area, we have to make sure that the function f(x) remains positive within the limits of integration.
Since f(0)=0, f(π/2)=0, and f(x) is positive between the two limits, we just have to evaluate the integral:
∫x(sin(x)-1)dx between 0 and π/2.
This can be done by integration by parts:
∫x(sin(x)-1)dx
= x(-cos(x)-x) + ∫1*(cos(x)+x)dx
= sin(x)+x²/2 - xcos(x) - x² + C
= sin(x) - xcos(x) - x²/2 + C
Evaluate the integral between the given limits (ignore the constant C).
I get (π²/8)-1.
    
f(x)=xsin(x)-x
=x(sin(x)-1)
between 0 and π/2.
Since we are looking for the area, we have to make sure that the function f(x) remains positive within the limits of integration.
Since f(0)=0, f(π/2)=0, and f(x) is positive between the two limits, we just have to evaluate the integral:
∫x(sin(x)-1)dx between 0 and π/2.
This can be done by integration by parts:
∫x(sin(x)-1)dx
= x(-cos(x)-x) + ∫1*(cos(x)+x)dx
= sin(x)+x²/2 - xcos(x) - x² + C
= sin(x) - xcos(x) - x²/2 + C
Evaluate the integral between the given limits (ignore the constant C).
I get (π²/8)-1.
                    Answered by
            MathMate
            
    Scott, if you cannot wait and repost, i suggest you indicate that you are reposting.  If teachers repeat the same work, the end result is everyone gets slower responses.
    
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