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Evaluate I = ∫ 1 0 ∫ √ 1 + X 2 0 D X . D Y 1 + X 2 + Y 2 - Applied Mathematics 2

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Sum

Evaluate I = `int_0^1 int_0^(sqrt(1+x^2)) (dx.dy)/(1+x^2+y^2)`

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Solution

I = `int_0^1 int_0^(sqrt(1+x^2)) (dx.dy)/(1+x^2+y^2)`

I= `int_0^1 1/(sqrt(1+x^2)) [tan^(-1)  y/sqrt(1+x^2)]_0^sqrt(1+x^2) dx` 

`therefore "I" = int_0^1 pi/4 1/sqrt(1+x^2) dx `

`therefore "I"=pi/4[log(x+sqrt(1+x^2))]_0^1`

`therefore  "I" =pi/4log(1+sqrt2)`

Concept: Double Integration‐Definition
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