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}
$

Comments

Popular posts from this blog

How does Bilbo show leadership and courage in The Hobbit?

What was the effect of World War II on African Americans?

How is Jack in William Golding's The Lord of the Flies presented as militaristic?