Question

`int_0^5 cos(π(x - [x/2]))dx` where [t] denotes greatest integer less than or equal to t, is equal to ______.

Options

• –3

• –2

• 2

• 0

Answer 1

`int_0^5 cos(π(x - [x/2]))dx` where [t] denotes greatest integer less than or equal to t, is equal to 0.

Explanation:

Let I = `int_0^5 cos(π(x - [x/2]))dx`

Now, `cos(π(x - [x/2])) = cos(πx - π[x/2])`

`cos(πx - π[x/2]) = {{:(cos(πx - 0), 0 < x < 2),(cos(πx - π), 2 ≤ x < 4),(cos(πx - 2π), 4 ≤ x < 5):}`

∴ I = `int_0^2 cos(πx)dx + int_2^4 cos(πx - π)dx + int_4^5 cos(πx - 2π)dx`

⇒ I = `[(sin(πx))/π]_0^2 + [(sin(πx - π))/π]_2^4 + [(sin(πx - 2π))/π]_4^5`

⇒ I = `1/π (sin2π - sin0 + sin(4π - π)) - sin(2π - π) + sin(5π - 2π) - sin(4π - 2π)`

⇒ I = `1/π(0)`

⇒ I = 0