$\int_{-4}^{4} \log \left(\frac{8-x}{8+x}\right) d x=$

$\int_{-4}^{4} \log \left(\frac{8-x}{8+x}\right) d x=$
  1. $-4$
  2. 8
  3. 4
  4. 0

Solution

$\begin{aligned} \text { Let } f(x) &=\log (8-x)-\log (8+x) \\ f(-x) &=\log (8+x)-\log (8-x) \\ &=-[\log (8-x)-\log (8+x)] \\ \text { Thus } f(-x) &=-f(x) \Rightarrow I=0 \end{aligned}$

Asked in: MHT CET 2020 (15 Oct Shift 2)

Practice more Definite Integration questions on Aicharya