Question

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

Answer 1

`\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.`