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F(x)=int_(2)^(2x)sqrt(15-t)dt

F^(')(x)=

$F(x)=\int_{2}^{2 x} \sqrt{15-t} d t$ $\newline$ $F^{\prime}(x)=$

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Evaluate definite integrals using the power rule

Full solution

Q.

F(x)=\int_{2}^{2 x} \sqrt{15-t} d t

\newline

F^{\prime}(x)=

Apply Fundamental Theorem of Calculus: Use the Fundamental Theorem of Calculus Part $1$ , which states that if $F(x) = \int_{a}^{g(x)}f(t)\,dt$ , then $F'(x) = f(g(x)) \cdot g'(x)$ .
Identify $f(t)$ and $g(x)$ : Identify $f(t)$ as $\sqrt{15-t}$ and $g(x)$ as $2x$ .
Calculate $g'(x)$ : Calculate $g'(x)$ , which is the derivative of $2x$ with respect to $x$ . $\newline$ $g'(x) = \frac{d}{dx}(2x) = 2$ .
Substitute into $F'(x)$ : Substitute $g(x) = 2x$ and $g'(x) = 2$ into the formula $F'(x) = f(g(x)) \cdot g'(x)$ . $\newline$ $F'(x) = \sqrt{15 - 2x} \cdot 2$ .
Simplify $F'(x)$ : Simplify the expression for $F'(x)$ . $F'(x) = 2 \cdot \sqrt{15 - 2x}$ .

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