Saved Bookmarks
| 1. |
Evaluate `int_(0)^(1)(tx+1-x)^(n)dx`, where n is a positive integer and t is a parameter independent of x . Hence , show that |
|
Answer» Correct Answer - `[(t^(n+1)-1)/((t-1)(n+1))]` Let ` I= int _(0)^(1) (tx =1-x)^(n) dx= int_(0)^(1){(t-1)x+1}^(n) dx` `=[(((t -1)x+1)^(n+1))/((n+1)(t-1))]_(0)^(1)=(1)/( n+1)((t^(n+1-1))/(t -1))` ` = (1)/(n+1)(1+t+t^(2)+ . . . + t^(n))` . . . (i) Again , `I= int_(0)^(1) ( tx +1-x)^(n) dx = int _(0)^(1) [ (1- x)+tx]^(n) dx` `=int _(0)^(1) [.^(n)C_(0)(1- x)^(n) + ^(n)C_(1)(1-x)^(n-1)(t x)` `=int_(0)^(1) [ sum_(r=0)^(n).^(n)C_(r)(1-x)^(n-r)( tx)^(r)]dx` ` = sum_(r =0)^(n) .^(n)C_(r)[int_(0)^(1)(1-x)^(n-r) .x^(r)dx]t^(r)` . . . (iii) From Eqs . (i) and (ii) , we get ` sum_(r =0)^(n) .^(n)C_(r)[int_(0)^(1)(1-x)^(n-r) .x^(r)dx]t^(r)=(1)/(n +1) (1+t+ . . . t^(n))` On equating coefficient of `t ^(k)` on both sides , we get ` .^(n)C_(k) [ int_(0)^(1) (1-x)^(n-k)*x^(k) dx ] =(1)/( n=1)` `rArr int_(0)^(1) (1- x)^(n-k)x^(k) dx = (1)/((n+1)^(n)C_(k)` |
|