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)=(\mathrm d)/(\mathrm dx)F(x)=(\mathrm d)/(\mathrm dx)\int_0^x√(t^3+1)\,\mathrm dt=√(x^3+1)`

which means


`F'(2)=√(2^3+1)=√(8+1)=\sqrt9=3`