#### To determine

**To prove:**

∫01fxdx=∫01f1-xdx

#### Answer

∫01fxdx=∫01f1-xdx

#### Explanation

**1) Concept:**

i) -∫abf(x)dx=∫baf(x)dx

ii) The substitution rule: If u=g(x) is a differentiable function whose range is I and f is continuous on I, then ∫f(gx)g'xdx=∫f(u)du. Here g(x) is substituted as u and g’(x)dx =du

**2) Given:**

∫01fxdx=∫01f1-xdx

**3) Calculation:**

Let us start with the right side of the equation

∫01f1-xdx

Substitute 1-x=u, therefore the differentiation gives -dx=du, dx=-du

And the limit also changes for x=0, u=1-0=1 and for x=1, u=1-1=0

Therefore, the given integral becomes

∫01f1-xdx=∫10fu-du

=-∫10fudu

=∫01fudu

=∫01fxdx

**Conclusion:**

Therefore, it is proved that ∫01fxdx=∫01f1-xdx