Question

Show that $\int_{0}^{a}$ƒ(x)g(x)dx=2$\int_{0}^{a}$ƒ(x)dx,if f and g are defined as ƒ(x)=ƒ(a-x)and g(x)+g(a-x)=4

Answer 1

Let I=$\int_{0}^{a}$ƒ(x)g(x)dx...(1)

⇒I=$\int_{0}^{a}$ƒ(a-x)g(a-x)dx) ($\int_{0}^{a}$ƒ(x)dx=$\int_{0}^{a}$ƒ(a-x)dx)

⇒I=$\int_{0}^{a}$ƒ(x)g(a-x)dx...(2)

Adding(1)and(2),we obtain

2I=$\int_{0}^{a}${ƒ(x)g(x)+ƒ(x)g(a-x)}dx

⇒2I=$\int_{0}^{a}$ƒ(x){g(x)+g(a-x)}dx

⇒2I=$\int_{0}^{a}$ƒ(x)×4dx [g(x)+g(a-x)=4]

⇒I=2$\int_{0}^{a}$ƒ(x)dx