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If f(0)=1,f(2)=3,f(2)=5,t h e nfin dt h ...

If `f(0)=1,f(2)=3,f(2)=5,t h e nfin dt h ev a l u eof` `int_0^1xf"(2x)dx

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`I_(1)=int_(0)^(1)xf''(2x)dx`. Putting `t=2x`, i.e.
`dx=(dt)/2`, we get
`I_(1)=1/4int_(0)^(2)tf''(t)dt`
`=1/4[tf'(t)|_(0)^(2)-int_(0)^(2)f'(t)dt]` (Integrating by parts)
`=1/4[tf'(t)|._(0)^(2)-f(t)|_(0)^(2)]`
`=1/4(2f'(2)-f(2)+f(0))`
`=1/4(10-3+1)=2`
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