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)

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

yea it is ∫ (t^3 + 2t^2 + 2) dt from 0 to x. thanks for ur help

To find the second derivative of the function f(x) = ∫(0,x) t^3 + 2t^2 + 2 dt, we need to first find the first derivative and then differentiate it once again.

Step 1: Find the first derivative, f'(x):
To find the derivative of f(x), we use the fundamental theorem of calculus, which states that if F(x) is the antiderivative of f(x), then the derivative of the definite integral from a to x of f(t) dt with respect to x is f(x).

In this case, f(x) = ∫(0,x) t^3 + 2t^2 + 2 dt. We can find F(x) by integrating the function t^3 + 2t^2 + 2 with respect to t.

∫(0,x) t^3 + 2t^2 + 2 dt = [t^4/4 + (2/3)t^3 + 2t] from 0 to x
= (x^4/4 + (2/3)x^3 + 2x) - (0^4/4 + (2/3)(0)^3 + 2(0))
= x^4/4 + (2/3)x^3 + 2x

So, F(x) = x^4/4 + (2/3)x^3 + 2x

Now, we can differentiate F(x) to find f'(x):
f'(x) = d/dx (x^4/4 + (2/3)x^3 + 2x)
= 4(x^3/4) + 3(2/3)x^2 + 2
= x^3 + 2x^2 + 2

Step 2: Find the second derivative, f''(x):
To find the second derivative, we differentiate f'(x) with respect to x.
f''(x) = d/dx (x^3 + 2x^2 + 2)
= 3x^2 + 4x

Therefore, the second derivative of the function f(x) is f''(x) = 3x^2 + 4x.