If F(x)=the integration from 0 to x of square root of (t^3 + 1)dt then what is F'(2)?

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Answer 1 · 2018-03-02T15:41:37

$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad F'(2) = 3.$

$"We are given:"$

$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ F(x) = int_0^x \ sqrt{ t^3 + 1 } \ dt.$

$"So, directly by a Fundamental Theorem of Calculus, we have"$
$"at once:"$

$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ F'(x) = sqrt{ x^3 + 1 }.$

$"Thus:"$

$\qquad \qquad \qquad \qquad \qquad \qquad F'(2) = sqrt{ 2^3 + 1 } = sqrt{ 9 } = 3.$

$"So:"$

$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad F'(2) = 3.$