Single Variable Calculus, Chapter 5, 5.3, Section 5.3, Problem 22

Find the integral $\displaystyle \int^1_{0} \left( 1 + \frac{1}{2} u^4 - \frac{2}{5} u^9 \right) du$

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 $\displaystyle f(u) = 1 + \frac{1}{2} u^4 - \frac{2}{5} u^9$, then


$
\begin{equation}
\begin{aligned}

F(u) =& 1 \left( \frac{u^{0 + 1}}{0 + 1} \right) + \frac{1}{2} \left( \frac{u^{4 + 1}}{4 + 1} \right) - \frac{2}{5} \left( \frac{u^{9 + 1}}{9 + 1} \right) + C
\\
\\
F(u) =& u + \frac{1}{2} \left( \frac{u^5}{5} \right) - \frac{2}{5} \left( \frac{u^{10}}{10} \right) + C
\\
\\
F(u) =& u + \frac{u^5}{10} - \frac{u^{10}}{25} + C

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

& \int^1_0 \left( 1 + \frac{1}{2} u^4 - \frac{2}{5} u^9 \right) du = F(1) - F(0)
\\
\\
& \int^1_0 \left( 1 + \frac{1}{2} u^4 - \frac{2}{5} u^9 \right) du =
1 + \frac{(1)^5}{10} - \frac{(1)^{10}}{25} + C - \left[ 0 + \frac{(0)^5}{10} - \frac{(0)^{10}}{25} + C \right]
\\
\\
& \int^1_0 \left( 1 + \frac{1}{2} u^4 - \frac{2}{5} u^9 \right) du = 1 + \frac{1}{10} - \frac{1}{25} + C - C
\\
\\
& \int^1_0 \left( 1 + \frac{1}{2} u^4 - \frac{2}{5} u^9 \right) du = \frac{53}{50}
\\
\\
& \text{ or }
\\
\\
& \int^1_0 \left( 1 + \frac{1}{2} u^4 - \frac{2}{5} u^9 \right) du = 1 . 06



\end{aligned}
\end{equation}
$