#### To determine

**To show:**

∫01xa1-xbdx=∫01xb1-xadx

#### Answer

∫01xa1-xbdx=∫01xb1-xadx

#### Explanation

**1) Concept:**

i. 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

ii.

∫abf(x)dx=-∫baf(x)dx

iii.

∫abf(u)du=∫abf(x)dx

**2) Given:**

f is continuous on R

**3) calculation:**

Let

∫01xa1-xbdx

Here use the substitution method

Substitute 1-x=u

Differentiating with respect to x

- dx=du

dx=-du

The limits changes; the new limits of integration are calculated by substituting

At x=0, u=1-0=1 and

At x=1, u=1-1=0

Therefore, the integral becomes

∫01xa1-xbdx=∫101-ua1-1-ub(-du)

Simplify

=∫10-1-uaubdu

Usingthe property of definite integral

=∫01ub1-uadu

Using concept iii)

=∫01xb1-xadx

Therefore,∫01xa1-xbdx=∫01xb1-xadx

**Conclusion:**

Therefore,∫01xa1-xbdx=∫01xb1-xadx