If f and g are defined as f(x) = f (a -x) and g(x) + g (a -x) = 4, then ∫0a f(x)g(x)dx is equal to
∫0a f(x)dx
2∫0a f(x)dx
2∫0a f(x)g(x)dx
None of these
Let,
I=∫0a f(x)g(x)dx----iI=∫0a f(a−x)g(a−x)dx∵∫0a f(x)dx=∫0a f(a−x)dx
⇒I=∫0a f(x){4−g(x)}dx---ii[∵f(x)=f(a−x) and g(x)+g(a−x)=4 (given) ]
On adding Eqs. (i) and (ir), we get
2I=∫0a 4f(x)dx⇒I=2∫0a f(x)dx