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