If the function `f : [0,8] to R` is differentiable, then for `0 < alpha <1 < beta < 2 , int_0^8 f(t) dt ` is equal to

Options

A. `3[alpha^(3)f(alpha^(2))+beta^(2)f(beta^(2))]`
B. `3[alpha^(3)f(alpha)+beta^(3)f(beta)]`
C. `3[alpha^(2)f(alpha^(3))+beta^(2)f(beta^(3))]`
D. `3[alpha^(2)f(alpha^(2))+beta^(2)f(beta^(2))]`

Answer

C

Explanation

Let `g(xS)=int_(0)^(x^(3))f(t)dt`
Now, `int_(0)^(8)f(t)dt=g(2)=(g(2)-g(1))/(2-1)+(g(1)-g(0))/(1-0)`
`g'(alpha)+g'(beta)`
`=3[alpha^(2)f(alpha^(3))+beta^(2)f(beta^(3))]`