Single Variable Calculus, Chapter 5, 5.3, Section 5.3, Problem 6

Find $\displaystyle g'(x)$ of $\displaystyle g(x) = \int^x_0 (1 + \sqrt{t}) dt$

a.) Using 1st Fundamental Theorem of Calculus

When $g(x) = \int^x_a f(t) dt, g'(x) = f(x)$

Let $x = t$ to $f(t)$, so we have

$g'(x) = 1 + \sqrt{x}$

b.) Evaluate the integral using 2nd Fundamental Theorem of Calculus.

$\displaystyle \int^b_a f(x) dx = F(b) - F(a)$ where $f$ is any anti-derivative of $f$. Such that $F' = f$


$
\begin{equation}
\begin{aligned}

& \frac{d}{dx} \left[ g(x) = \int^x_0 (1 + \sqrt{t}) dt = F(x) - F(0) \right] \frac{d}{dx}
&& \text{By differentiating both sides}
\\
\\
& g'(x) = F'(x) - F'(0), \text{ So}
\\
\\
& g'(x) = f(x) - f(0)
&& \text{Since } \displaystyle \int^a_a \text{ of any function is so $f(0) = 0$, then}
\\
\\
& g'(x) = 1 + \sqrt{x}

\end{aligned}
\end{equation}
$

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