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: Substitution:
Let \(u = x + y\). Then \(\frac{du}{dx} = 1 + \frac{dy}{dx} = 1 + \sin u + \cos u\). Separate variables: \(\frac{du}{1 + \sin u + \cos u} = dx\).

Step 2: Half angle substitution:
Let \(t = \tan\frac u2\). Then \(\sin u = \frac{2t}{1 + t^2}\), \(\cos u = \frac{1 - t^2}{1 + t^2}\) and \(du = \frac{2\,dt}{1 + t^2}\).
\[ 1 + \sin u + \cos u = \frac{(1 + t^2) + 2t + (1 - t^2)}{1 + t^2} = \frac{2(1 + t)}{1 + t^2} \]

Step 3: Integrate:
\[ \int\frac{2\,dt/(1 + t^2)}{2(1 + t)/(1 + t^2)} = \int\frac{dt}{1 + t} = \log(1 + t) = x + c \]
So \(\log\left[1 + \tan\frac{x+y}{2}\right] = x + c\). The right hand side has x (not y), because we integrated with respect to x, which matches option (C).

Final Answer:
The solution is \(\log\left[1 + \tan\frac{x+y}{2}\right] = x + c\), option (C). \[ \boxed{\log\left[1 + \tan\frac{x+y}{2}\right] = x + c} \]