If [x] is the greatest integer x, then the value of the integral _-0.9^0.9([x^2]+(2-x2+x))…

Question

If $[x]$ is the greatest integer $\leq x$, then the value of the integral $\displaystyle\int_{-0.9}^{0.9}\left([x^2]+\log\left(\dfrac{2-x}{2+x}\right)\right)dx$ is:

Accepted Answer

$\int_{-0.9}^{0.9}[x^2]\,dx + \int_{-0.9}^{0.9}\log\left(\dfrac{2-x}{2+x}\right)dx = 0 + \int_{-0.9}^{0.9}\log\left(\dfrac{2-x}{2+x}\right)dx$

(since $[x^2]=0$ for $-0.9 \leq x \leq 0.9$, being an even function, its integral doesn't vanish in general but here $\int [x^2]dx=0$ since $x^2<1$ throughout)

Let $f(x) = \log\left(\dfrac{2-x}{2+x}\right)$; $f(-x) = \log\left(\dfrac{2+x}{2-x}\right) = -f(x)$, so $f$ is odd.

Hence the required integral $=0$

Solutions · 1