Single Variable Calculus, Chapter 5, 5.2, Section 5.2, Problem 44

Evaluate $\displaystyle \int^5_2 \left( 1 + 3x^4 \right)dx$ using properties of integrals.
Using properties of integral

$
\begin{equation}
\begin{aligned}
\int^b_a [ f(x) + g(x) ] dx &= \int^b_a f(x) dx + \int^b_a g(x) dx\\
\\
\int^5_2 (1 + 3x^4) dx &= \int^5_2 1 dx + \int^5_2 3x^4 dx, \text{ then}
\end{aligned}
\end{equation}
$

Apply $\displaystyle \int^b_a c d x = c(b-a) \text{ and } \int^b_a cf(x) dx = c \int^b_a f(x) dx$, where $c$ is any constant so we have,
$\displaystyle 1(5-2)+3 \int^5_2 x^4 dx = 3 + 3 \int^5_2 x^4 dx \qquad \Longleftarrow \text{(Equation 1)}$
Solving for $\displaystyle \int^5_2 x^4 dx$

Here we have $f(x) = x^4$, $a = 2$, $b = 5$ and

$
\begin{equation}
\begin{aligned}
\Delta x &= \frac{b-a}{n}\\
\\
\Delta x &= \frac{5-2}{n}\\
\\
\Delta x &= \frac{3}{n}
\end{aligned}
\end{equation}
$


$
\begin{equation}
\begin{aligned}
x_i &= a + i \Delta x\\
\\
x_i &= 2 + \frac{3i}{n}
\end{aligned}
\end{equation}
$


Using defintion of integral

$
\begin{equation}
\begin{aligned}
\int^a_b f(x) dx &= \lim_{n \to \infty} \sum\limits_{i = 1}^n f(x_i) \Delta x\\
\\
\int^5_2 x^4 dx &= \lim_{n \to \infty} \sum\limits_{i = 1}^n f\left(2+\frac{3i}{n} \right) \left( \frac{3}{n} \right)\\
\\
\int^5_2 x^4 dx &= \lim_{n \to \infty} \sum\limits_{i = 1}^n \left( 2 + \frac{3i}{n} \right)^4 \left( \frac{3}{n} \right)\\
\\
\int^5_2 x^4 dx &= \lim_{n \to \infty} \sum\limits_{i = 1}^n \left[ 16 +32 \left(\frac{3i}{n} \right) + 24 \left(\frac{3i}{n} \right)^2 + 8 \left(\frac{3i}{n} \right)^3 + \left( \frac{3i}{n} \right)^4 \right] \left( \frac{3}{n} \right)\\
\\
\int^5_2 x^4 dx &= \lim_{n \to \infty} \sum\limits_{i = 1}^n \left( 16 + \frac{96i}{n} + \frac{216i^2}{n^2} + \frac{216i^3}{n^3} + \frac{81 i^4}{n^4} \right) \left( \frac{3}{n} \right)\\
\\
\int^5_2 x^4 dx &= \lim_{n \to \infty} \sum\limits_{i = 1}^n \left( \frac{48}{n} + \frac{288 i}{n^2} + \frac{648i^2}{n^3} + \frac{648i^3}{n^4} + \frac{243 i^4}{n^5} \right)\\
\\
\int^5_2 x^4 dx &= \lim_{n \to \infty} \left( \frac{1}{n} \sum\limits_{i = 1}^n 48 + \frac{288}{n^2} \sum\limits_{i = 1}^n i + \frac{648}{n^3} \sum\limits_{i = 1}^n i^2 + \frac{648}{n^4} \sum\limits_{i = 1}^n i^3 + \frac{243}{n^5} \sum\limits_{i = 1}^n i^4 \right)\\
\\
\int^5_2 x^4 dx &= \lim_{n \to \infty} \left[ \frac{1}{\cancel{n}} (48\cancel{n}) + \frac{288}{n^2} \left( \frac{n(n+1)}{2} \right) + \frac{648}{n^3} \left( \frac{n(n+1)(2n+1)}{6} \right) + \frac{648}{n^4} \left( \frac{n(n+1)}{2} \right)^2 + \frac{243}{n^5} \left( \frac{n(n+1)(2n+1)(3n^2+3n-1)}{3} \right) \right]\\
\\
\int^5_2 x^4 dx &= \lim_{n \to \infty} \left[ 48 + \frac{144(n+1)}{n} + \frac{108(n+1)(2n+1)}{n^2} + \frac{162(n+1)^2}{n^2} + \frac{81(n+1)(2n+1)(3n^2+3n-1)}{n^4} \right]\\
\\
\int^5_2 x^4 dx &= \lim_{n \to \infty} \left[ 48 + \frac{144n+144}{n} + \frac{108(2n^2+3n+1)}{n^2} + \frac{162(n^2+2n+1)}{n^2} + \frac{81(2n^2+3n+1)(3n^2+3n-1)}{n^4} \right]\\
\\
\int^5_2 x^4 dx &= \lim_{n \to \infty} \left[ 48 + \frac{144n+144}{n} + \frac{216n^2+324n+108+162n^2+324n+162}{n^2} + \frac{81\left( 6n^4 + 6n^3 - 2n^2 + 9n^3 + 9n^2 - \cancel{3n} + 3n^2 + \cancel{3n} - 1 \right)}{n^4}\right]\\
\\
\int^5_2 x^4 dx &= \lim_{n \to \infty} \left[ 48 + \frac{144n+144}{n} + \frac{378 n^2 + 648n + 170}{n^2} + \frac{81\left(6n^4 +15n + 10n^2 -1\right)}{n^4}\right]\\
\\
\int^5_2 x^4 dx &= \lim_{n \to \infty} \left( 48 + \frac{144n+144}{n} + \frac{378n^2 + 648n + 170}{n^2}+\frac{486n^4 +1215n^3+810n^2-81}{n^4}\right)\\
\\
\int^5_2 x^4 dx &= \lim_{n \to \infty} \left( \frac{48n^4 + 144n^4 + 144n^3 + 378n^4 + 648n^3 + 170n^2 + 486n^4 + 1215n^3 + 810n^2 - 81}{n^4} \right)\\
\\
\int^5_2 x^4 dx &= \lim_{n \to \infty} \left( \frac{1056n^4 + 2007 n^3 + 980n^2 - 81}{n^4} \right)\\
\\
\int^5_2 x^4 dx &= \lim_{n \to \infty} \left( \frac{\frac{1056\cancel{n^4}}{\cancel{n^4}} + \frac{2007n^3}{n^4} + \frac{980n^2}{n^4} - \frac{81}{n^4}}{\frac{\cancel{n^4}}{\cancel{n^4}} } \right)\\
\\
\int^5_2 x^4 dx &= \lim_{n \to \infty} \left( 1056 + \frac{2007}{n} + \frac{980}{n^2} - \frac{81}{n^4} \right)\\
\\
\int^5_2 x^4 dx &= 1056 + \lim_{n \to \infty} \frac{2007}{n} + \lim_{n \to \infty} \frac{980}{n^2} - \lim_{n \to \infty} \frac{81}{n^4}\\
\\
\int^5_2 x^4 dx &= 1056 +0 +0 -0\\
\\
\int^5_2 x^4 dx &= 1056
\end{aligned}
\end{equation}
$


Since Equation 1 is

$
\begin{equation}
\begin{aligned}
3 + 3 \int^5_2 x^4 dx &= 3+3 (1056)\\
\\
3 + 3 \int^5_2 x^4 dx &= 3 + 3168\\
\\
3 + 3 \int^5_2 x^4 dx &= 3171
\end{aligned}
\end{equation}
$