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${:[F(x)=int_(0)^(sqrtx)t^(2)dt],[F^(')(x)=]:}$

$\begin{array}{l}F(x)=\int_{0}^{\sqrt{x}} t^{2} d t \\ F^{\prime}(x)=\end{array}$

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Evaluate definite integrals using the chain rule

Full solution

Q.

\begin{array}{l}F(x)=\int_{0}^{\sqrt{x}} t^{2} d t \\ F^{\prime}(x)=\end{array}

Apply Fundamental Theorem of Calculus: We need to use the Fundamental Theorem of Calculus to find $F'(x)$ , which tells us that if $F(x)$ is defined as an integral from $a$ to $x$ of $f(t)\,dt$ , then $F'(x) = f(x)$ .
Define $F(x)$ and $f(t)$ : $F(x) = \int_{0}^{\sqrt{x}}t^{2}\,dt$ , so $f(t) = t^2$ . To find $F'(x)$ , we replace $t$ with $\sqrt{x}$ in $f(t)$ .
Calculate $F'(x)$ : $F'(x) = (\sqrt{x})^2 = x$ .

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