Posted by MILY on .
the fundamental theorem of calculus,
f(x)=∫(0,x) t^3+2t^2+ 2 dt,
and find f"(x).
my answer was:
f'(x)=x^3+x^2
f"(x)=3x^2+2x
it said its wrong. i don't know how solve with interval, ∫(0,x)

COLLEGE CALCULUS. HELP! 
Reiny,
I must admit I have never seen that notation before
is this ∫ (t^3 + 2t^2 + 2) dt from 0 to x ??
if so,
= (1/4)t^4 + (2/3)t^3 + 2t  from 0 to x
= (1/4)x^4 + (2/3)x^3 + 2x
so f(x) = (1/4)x^4 + (2/3)x^3 + 2x
f'(x) = x^3 + 2x^2 + 2  as expected, the original
f''(x) = 3x^2 + 4x 
COLLEGE CALCULUS. HELP! 
MILY,
yea it is ∫ (t^3 + 2t^2 + 2) dt from 0 to x. thanks for ur help