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

1 Answer
Mar 2, 2018

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

Explanation:

"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.