Saved Bookmarks
| 1. |
If x=phi(t) is a differentiable function of 't', then prove thatintf(x)dx=intf[phi(t)]phi'(t)dt. |
|
Answer» Solution :`x = phi(t)` is differentiable function of t. `THEREFORE""(dy)/(dt)=phi'(t)` `"LET"intf(x)dx=F(x)` `therefore""(dx)/(dt)[F(x)]=f(x)` `(d)/(dt)[F(x)]=(d)/(dx)[F(x)].(dx)/(dt)` `"(Using chain rule)"` `=f(x).(dx)/(dt)` `=f[phi(t)].phi'(t)` `therefore""` By the DEFINITION of integral, `F(x)=INT f[phi(t)].phi'(t)dt` `therefore""intf(x)dx=intf[phi(t)].phi'(t) dt.` Hence PROVED. |
|