Question

Question #53b73

Answer 1

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)`