Single Variable Calculus, Chapter 5, 5.3, Section 5.3, Problem 24

Find the integral $\displaystyle \int^8_1 \sqrt[3]{x} dx$

Using 2nd Fundamental Theorem of Calculus

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

Let $\displaystyle f(x) = \sqrt[3]{x}$ or $f(x) = (x)^{\frac{1}{3}}$, then


$
\begin{equation}
\begin{aligned}

F(x) =& \frac{x^{\frac{1}{3}+1}}{\displaystyle \frac{1}{3} + 1} + C
\\
\\
F(x) =& \frac{x^{\frac{4}{3}}}{\displaystyle \frac{4}{3}} + C
\\
\\
F(x) =& \frac{3x^{\frac{4}{3}}}{4} + C

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

& \int^8_1 \sqrt[3]{x} dx = F(8) - F(1)
\\
\\
& \int^8_1 \sqrt[3]{x} dx = \frac{3(8)^{\frac{4}{3}}}{4} + C - \left[ \frac{3(1)^{\frac{4}{3}}}{4} + C \right]
\\
\\
& \int^8_1 \sqrt[3]{x} dx = \frac{3 [(8)^{\frac{1}{3}}]^4}{4} + C - \frac{3}{4} - C
\\
\\
& \int^8_1 \sqrt[3]{x} dx = \frac{3(2)^4}{4} - \frac{3}{4}
\\
\\
& \int^8_1 \sqrt[3]{x} dx = \frac{3(16) - 3}{4}
\\
\\
&\int^8_1 \sqrt[3]{x} dx = \frac{48 - 3}{4}
\\
\\
& \int^8_1 \sqrt[3]{x} dx = \frac{45}{4}
\\
\\
& \text{ or }
\\
\\
& \int^8_1 \sqrt[3]{x} dx = 11.25

\end{aligned}
\end{equation}
$