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Question #53b73

1 Answer

Nov 27, 2016

I get #1-sqrt2/3#

Explanation:

On #(0,1)#, #{x^2} = x^2#

On #(1,sqrt2)#, #{x^2} = x^2-1#

To evaluate #int_0^sqrt2 {x^2} dx#, evaluate both and add:

#int_0^1 x^2 dx + int_1^sqrt2 (x^2-1) dx = (1/3) + (2/3-sqrt2/3)#

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