Let f:R→R and g:R→R be continuous functions, then the value of ∫π2−π2(f(x)+f(−x))(g(x)−g(−x))dx, is equal to
A
−1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
0
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
none of these
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is C0 Let I=∫π2−π2(f(x)+f(−x))(g(x)−g(−x))dx ...(1) Using property ∫baf(x)dx=∫baf(a+b−x)dx I=∫π2−π2(f(−x)+f(x))(g(−x)−g(x))dx ...(2) Adding (1) and (2), we get 2I=∫π2−π2(f(−x)+f(x))((g(x)−g(−x))+(g(−x)−g(x)))dx=0⇒I=0