Question

If F(x)= the integral from (0,x) the squareroot of (t^3+1)dt then F' (2)

Answer 1

\[\int\limits_{0}^{x}\sqrt{t ^{3}+1}\]

Answer 2

might be better like this ...

\[F(t)=\int F'(t)dt\]
\[\int_{0}^{x}F'(t)dt=F(x)-F(0)\]
\[\frac{d}{dx}F(x)-F(0)=F'(x)x' - F'(0)0'\]

Answer 3

so my guess is; sqrt(2^3 +1) = sqrt(9) = 3

Answer 4

thank you

Answer 5

yw