Question

If `int_0^1(sqrt(2x) - sqrt(2x - x^2))dx = int_0^1(1 - sqrt(1 - y^2) - y^2/2)dy + int_1^2(2 - y^2/2)dy` + I then I equal.

Options

• `int_0^1(1 + sqrt(1 - y^2))dy`

• `int_0^1(y^2/2 - sqrt(1 - y^2) + 1)dy`

• `int_0^1(1 - sqrt(1 - y^2))dy`

• `int_0^1(y^2/2 + sqrt(1 - y^2) + 1)dy`

Answer 1

`bb(int_0^1(1 - sqrt(1 - y^2))dy)`

Explanation:

`int_0^2(sqrt(2x) - sqrt(2x - x^2))dx = int_0^1(1 - sqrt(1 - y^2) - y^2/2)dy + int_1^2(2 - y^2/2)dy + I`

Now, L.H.S. = `int_0^2 sqrt(2x)dx - int_0^2sqrt(2x - x^2)dx`

= `sqrt(2)[2/3x^(3/2)]_0^2 - int_0^2 sqrt(1 - (x - 1)^2)dx`

= `sqrt(2)[(4sqrt(2))/3 - 0] - 2int_0^1 sqrt(1 - y^2)dy`

= `8/3 - 2int_0^1 sqrt(1 - y^2)dy`

And R.H.S. = `int_0^1(1 - y^2/2)dy - int_0^1sqrt(1 - y^2)dy + int_1^2(2 - y^2/2)dy + I`

= `[y - y^3/6]_0^1 - int_0^1sqrt(1 - y^2)dy + [2y - y^3/6]_1^2 + I`

= `[1 - 1/6] - int_0^1sqrt(1 - y^2)dy + [4 - 8/6 - 2 + 1/6] + I`

= `5/6 - int_0^1sqrt(1 - y^2)dy + 5/6 + I`

= `5/3 - int_0^1sqrt(1 - y^2)dy + I`

So, `8/3 - 2int_0^1sqrt(1 - y^2)dy = 5/3 - int_0^1sqrt(1 - y^2)dy + I`

⇒ I = `1 - int_0^1sqrt(1 - y^2)dy`

⇒ I = `int_0^1(1 - sqrt(1 - y^2))dy`