Definite integral graph ( finding area)

Question

I need to solve this integral using geometry. But my problem is graphing . I dont know on the graph which parts to shade in and evaluate. 

 

₀∫^3/2 sqrt(2x-x^2) 

 

After completing the square I get (x-1)^2+y^2= 1 . On the graph though I drew a semi circle from 0 to 3/2 but now I dont know what to do . Is the area 1/2(pi)(1^2) ?

Answer 1

2x-x² = -(x²-2x)

 

        = -(x²-2x+1) +1 = 1-(x-1)²

 

So, ∫_{(from 0 to 1.5)}√(2x-x²)dx = ∫_{(from 0 to 1.5)}√(1-(x-1)²)dx

 

                                           Let x-1= sinθ  then dx = cosθdθ

                                           When x = 0, sinθ = -1.  So, θ=-π/2

                                           When x = 1.5, sinθ = 1/2.  So, θ = π/6

 

We have ∫_{(from θ=-π/2 to θ=π/6)} √(1-sin²θ)cosθdθ

 

             = ∫_{(from -π/2 to π/6)}cos²θdθ

 

             = ∫_{(from -π/2 to π/6)}[(1+cos(2θ))/2]dθ

 

             = (1/2)[θ + (1/2)sin(2θ)]_{(from -π/2 to π/6)}

 

             = [π/12 + (1/4)sin(π/3)] - [-π/4 + (1/4)sin(-π)]

 

             = π/12 + π/4 + √3/8

 

             = π/3 + √3/8 ≈ 1.26