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f:[0,5]rarrR,y=f(x) such that f''(x)=f''(5-x)AAx in [0,5] f'(0)=1 and f'(5)=7, then the value of int_(1)^(4)f'(x)dx is

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Solution :`int_(1)^(4)f'(X)dx=[xf'(x)]_(1)^(4)-int_(1)^(4)xf''(x)`
Now `I=int_(1)^(4)xf''(x)dx=int_(1)^(4)(5-x)f''(5-x)dx`
`=5int_(1)^(4)f''(x)dx=-I`
`therefore""I=(5)/(2)[f'(4)-f'(1)]`
`therefore""int_(1)^(4)f'(x)dx=(3)/(2)[f'(4)+f'(1)]`
Now, `f''(x)=f''(5-x)`
`rArr""f'(x)=-f'(5-x)+C`
`rArr""f'(0)+f'(5)=c rArr c=8`
`"so"f'(x)+f'(5-x)=8 rArr f'(4)+f'(1)=8`


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