Let f:(0, ) and F(x)= 0x tf(t) dt. If F (x2 )=x4+x5, then r=112f (r2 ) is equal to .
Answer
219
Solution
F (x2 )=x4+x5 (t)=t2+t5/2
'(t)=2t+ 52t3/2=t f(t)
( Newton–Leibnitz rule)
f(t)=2+ 52 t f (r2 )=2+ 52r
r=112f (r2 )=24+ 52 12 132=24+195=219
Answer: 219