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September 2, 2014

September 2, 2014

Posted by **MILY** on Tuesday, April 17, 2012 at 2:36pm.

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 dont know how solve with interval, ∫(0,x)

- COLLEGE CALCULUS. HELP! -
**Reiny**, Tuesday, April 17, 2012 at 3:25pmI 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**, Tuesday, April 17, 2012 at 4:10pmyea it is ∫ (t^3 + 2t^2 + 2) dt from 0 to x. thanks for ur help

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