Using Properties
∫2−ag(x)dx=2∫a0g(x)dx ...(1)
And ∫bag(x)dx=∫bag(a+b−x)dx ...(2)
A) Using (1) ∫a−a(f(x))2dx=2∫a0(f(x))2dx
B) Let I=∫a−ax(f(x)2dx ...(3)
From (2) I=−∫a−ax(f(x)2dx (4)
From (3) and (4) we get 2I=0⇒I=0
C) Let J=∫a−a(x2+x3)(f(x))2dx ...(5)
From (2) J=∫a−a(x2−x3)(f(x))2dx ...(6)
From (5) and (6)
2J=∫a−ax2(f(x))2dx
Now using (1) 2J=2∫a0x2(f(x))2dx⇒J=∫a0x2(f(x))2dx
D) Let K=∫a−a(f(x))4dx=2∫a0(f(x))4dx from (1)
And from (2) K=2∫a0(f(a−x))4dx
Therefore
2K=2∫a0(f(x))4dx+2∫a0(f(a−x))4dx⇒K=∫a0(f(x))4dx+∫a0(f(a−x))4dx