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definite integral 3 to 6 square root of (x+square root of 12x-36) plus square root of (x-square root of 12x-36) dx = ?
Dear
I= 3∫6 {√[x+√(12x-36)] + √[x-√(12x-36)] }dx
or
I = 3∫6 {√[9-x+√(72-12x)] + √[9-x-√(72-12x)] }dx
=3∫6 {√[3 +(6-x)+2√3√(6-x)] + √[3+(6-x)-2√3√(6-x)] }dx
=3∫6 {|√3+√(6-x)| + |√3 - √(6-x)| }dx
=3∫6 {√3+√(6-x) + √3 - √(6-x) }dx
= 3∫6 2√3dx
=6√3
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