Intergal question

[ringyal1]

If f is a continuous function, show that definite integral 0to 1 of f(x)dx = integral 0 to 1 of f(1-x)dx

Kindly help me solve this problems thanks...

 

[Deleted member 4993]

If f is a continuous function, show that definite integral 0to 1 of f(x)dx = integral 0 to 1 of f(1-x)dx

Kindly help me solve this problems thanks...

Are you sure you have the limits correct?

 

[BigGlenntheHeavy]

$\displaystyle Prove \ \int_{0}^{1}f(x)dx \ = \ \int_{0}^{1}f(1-x)dx$

$\displaystyle \int_{0}^{1}f(x)dx \ = \ F(1)-F(0), \ FTC \ I$

$\displaystyle \int_{0}^{1}f(1-x)dx, \ let \ u \ = \ 1-x, \ then \ du \ = \ -dx \ \implies \ -du \ = \ dx$

$\displaystyle \int_{0}^{1}f(1-x)dx \ = \ -\int_{1}^{0}f(u)du \ = \ \int_{0}^{1}f(u)du \ = \ F(1)-F(0)$

$\displaystyle Hence, \ \int_{0}^{1}f(x)dx \ = \ \int_{0}^{1}f(1-x)dx, \ QED$

 

[ringyal1]

Thank u BigGlenntheHeavy.

Can you explain when the limits inverse when we substitute -du=dx. dont we need one more -ve sign to inverse the limits?

 

[BigGlenntheHeavy]

Let u =1-x, then where ever x = zero, u =1 and where ever x =1, u =0.