Let f : R → R and g : R → R be continuous functions, then the value of the integral ∫π2−π2[f(x)+f(−x)][g(x)−g(−x)]dx= [IIT 1990; DCE 2000; MP PET 2001]
A
π
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
-1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
\N
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution
The correct option is D \N Leth(x)={f(x)+f(−x)}{g(x)−g(−x)} h(−x)={f(−x)+f(x)}{g(−x)−g(x)} =−{f(−x)+f(x)}{g(x)−g(−x)}=−h(x) Therefore,∫π2−π2h(x)dx=0