This can be written ∫[3,7]((∜(4-(x-5)²)dx) or 2∫[3,5]((∜(4-(x-5)²)dx) (because of symmetry around x=5) where square brackets denote the high and low limits for the definite integration. Let x-5=2sinθ, then dx=2cosθdθ and limits for θ become [-π/2,π/2].
[(x-3)(7-x)=-x²+10x-21=-x²+10x-21=4-x²+10x-25=4-(x-5)². When x=3, 2sinθ=-2, θ=-π/2, when x=7, 2sinθ=2, θ=π/2. ∜(4-(x-5)²)=∜(4cos²θ)=√(2cosθ) and we substitute for dx.]
We now have the integral 2√2∫[-π/2,π/2]((cosθ)^(3/2)dθ).
The curve for the integrand is symmetrical about the vertical axis so the integral can be written: 4√2∫[0,π/2]((cosθ)^(3/2)dθ).
I don’t know how to integrate this function, so since we have a definite integral I have treated the integration as the area under a curve. This area I divided into narrow rectangles of height f(θ) (the function in the integrand) and width ∆θ. The sum of the areas is an approximation of the integral. I have shown some of the calculations below:
I chose ∆θ=0.01. The integral has the approximate value 4√2×0.879=4.97. A closer approximation to 2 decimal places is 4.95. The same result is achieved using the original integral.