Question

For a real number x let $\left[x\right]$ denote the largest integer less than or equal to x and $\left\{x\right\}=x-\left[x\right]$. Let n be a positive integer. Then ${\int }_0^n\cos \left(2\pi \right[x\left]\right\{x\left\}\right)dx$ is equal to.

• A. $0$
• B. $1$
• C. n
• D. $2n-1$

Answer 1

Answer: (A)

$0$

${\int }_0^n\cos \left(2\pi \right[x\left]\right\{x\left\}\right)dx$

$={\int }_0^1\cos (2\pi \times 0)dx+{\int }_1^2\cos \left(2\pi \right(x-1\left)\right)dx+...+{\int }_{n-1}^n\cos (2\pi \times (n-1\left)\right(x-n+1\left)\right)dx$

$=0+{\left[\frac{\sin \left(2\pi \right(x-1\left)\right)}{2\pi }\right]}_1^2+...+{\left[\frac{\sin \left(2\pi \right(n-1\left)\right(x-n+1\left)\right)}{2\pi (n-1)}\right]}_{n-1}^n$

Since $\sin \left(2n\pi \right)=0$ when $n$ is an integer, the integration sums up to zero.