Question

With the usual notation `int_1^2 ([x^2] - [x]^2)dx` is equal to ______.

Options

• `4 + sqrt(2) - sqrt(3)`

• `4 - sqrt(2) + sqrt(3)`

• `4 - sqrt(2) - sqrt(3)`

• none of these

Answer 1

With the usual notation `int_1^2 ([x^2] - [x]^2)dx` is equal to `underlinebb(4 - sqrt(2) - sqrt(3))`.

Explanation:

I = `int_1^2[x^2]dx - int_1^2[x]^2dx`

= `int_1^sqrt(2)dx + int_sqrt(2)^sqrt(3)2dx + int_sqrt(3)^2 3dx - int_1^2 1dx`

= `4 - sqrt(2) - sqrt(3)`