Question

Math

Posted 6 months ago

`$\begin{array}{l}g(x)=\int_{1}^{x} \sqrt{2 t+7} d t \\ g^{\prime}(9)=\end{array}$`

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Answer from Sia

Posted 6 months ago

Solution by Steps

step 1

Apply the Fundamental Theorem of Calculus Part 1 to differentiate the integral

step 2

According to the theorem, the derivative of the integral from a constant to x of a function is the function evaluated at x

step 3

So, $g'(x) = \frac{d}{dx} \int_{1}^{x} \sqrt{2t+7} \, dt = \sqrt{2x+7}$

step 4

Evaluate the derivative at $x=9$: $g'(9) = \sqrt{2(9)+7}$

step 5

Simplify the expression: $g'(9) = \sqrt{18+7} = \sqrt{25}$

step 6

Calculate the final value: $g'(9) = 5$

Answer

$g'(9) = 5$

Key Concept

Fundamental Theorem of Calculus Part 1

Explanation

The derivative of the integral of a function from a constant to x is the function evaluated at x.

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