Question

If $I={\int }_{-1}^1\left(\right[x^2]+\log \left(\frac{2+x}{2-x}\right))dx\left(1\right)$ where $\left[x\right]$ denotes the greatest integer $\le x,$ then $I$ equals

• A. $-2$
• B. $-1$
• C. $0$
• D. $1$

Answer 1

The correct option is C $0$

As $\log \left(\frac{2+x}{2-x}\right)$ is an odd function, we can write $I={\int }_{-1}^1\left[x^2\right]dx+0$
But for $-1<x<1,0\le x^2<1$ and thus,$\left[x^2\right]=0\therefore I=0.$
Ans: C