Question

Let \(f(x) = x-[x]\) for every real number \(x\), where \([x]\) is integral part of \(x\). then \(\int _{-1}^1f(x)\,dx\) is

• A. \(1\)
• B. \(2\)
• C. \(1/2\)
• D. \(0\)

Hint:

Substitute u = x + y and use the half angle tangent substitution.

Answer 1

The Correct Option is A

Step 1: Factor the denominator:
$1 + \sin u + \cos u = 2\cos\frac u2\left(\sin\frac u2 + \cos\frac u2\right)$, using $1 + \cos u = 2\cos^2\frac u2$ and $\sin u = 2\sin\frac u2\cos\frac u2$.

Step 2: Integrate:
\[ \int\frac{du}{2\cos\frac u2\left(\sin\frac u2 + \cos\frac u2\right)} = \int\frac{\frac12\sec^2\frac u2\,du}{1 + \tan\frac u2} = \log\left(1 + \tan\frac u2\right) \]
after dividing top and bottom by $\cos^2\frac u2$. This equals $x + c$. Option (C).

Final Answer:
Option (C). \[ \boxed{\text{(C)}} \]