02-03-2017
Mathematics

contestada

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

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LammettHash LammettHash

02-03-2017

By the fundamental theorem of calculus,

[tex]\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}[/tex]

which means

[tex]F'(2)=\sqrt{2^3+1}=\sqrt{8+1}=\sqrt9=3[/tex]

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