Question

If F(x)= integral from 0 to x of square root of (t^3 +1), then F'(2)=

Answer 1

By the fundamental theorem of calculus,

$\displaystyle F'(x)=\frac{\mathrm d}{\mathrm dx}F(x)=\frac{\mathrm d}{\mathrm dx}\int_0^x\sqrt{t^3+1}\,\mathrm dt=\sqrt{x^3+1}$

which means

$F'(2)=\sqrt{2^3+1}=\sqrt{8+1}=\sqrt9=3$