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.


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