Let `f:(0,prop) to R` be continous function such that
F(x)`=int_(0)^(x) t f(t) dt`. If `F(x^(2))=x^(4)+x^(5)`, then `sum_(r=1)^(12) f(r^(2))` is equal to

Options

A. 216
B. 219
C. 222
D. 225

Answer

B

Explanation

We have,
`F(x)=underset(0)overset(x)int t f(t) dt`
`rArr F(x^(2))=underset(0)overset(x^(2))int t f(t)dt`
`rArr x^(4)+x^(5)=undreset(0)overset(x^(2))int t f(t) dt " "[:' F(X^(2))=x^(4)+x^(5)]`
Differentiating both sides with respect to x, we get
`4x^(3)+5x^(4)underset(0)overset(x^(2)) 0 dt +(d)/(dx(x^(2))xx x^(2)f(x^(2))-0`
`rArr 4x^(3)+5x^(4)=2x^(3)f(x^(2))`
`rArr f(x^(2))=2+(5)(2)x`
`f(r^(2))=2+(5)/(2)r`
`:. underset(r=1)overset(12)sumf(r^(2))=underset(r=1)overset(12)sum(2+(5)/(2)r)=2xx12+(5)/(2)xx(12)/(2)(12+1)=219`