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\documentclass[12pt]{article}
\usepackage[margin=1in]{geometry} 
\usepackage{amsmath,amsthm,amssymb,amsfonts}

\newenvironment{problem}[2][]{
    \begin{trivlist}
        \item[
            {\bfseries #1}
            {\bfseries #2.}
        ]
}{\end{trivlist}}

\pagenumbering{gobble}

\newcommand{\assignment}{Pg. 423 \#35-57 odd, 73, 75, 93, 99}
\newcommand{\name}{<NAME>}
\newcommand{\duedate}{2019-02-25}
\newcommand{\details}{\textbf{\\\name \\\duedate \\\assignment}}

\begin{document}

\details

\begin{problem}{35}
$\displaystyle\int_0^{1.5} (x - x^2) \, dx$
\begin{align}
&= \dfrac{x^2}{2} - \dfrac{x^3}{3} \\
&= \left(\dfrac{(1.5)^2}{2} - \dfrac{(1.5)^3}{3}\right) + \left(\dfrac{(0)^2}{2} - \dfrac{(0)^3}{3}\right) \\
&= 0; \quad \text{The area above the x-axis is equal to the area below the x-axis}
\end{align}
\end{problem}

\begin{problem}{37}
$\displaystyle\int_{-1}^1 (x^4 - x^2) \, dx$
\begin{align}
&= \dfrac{x^5}{5} - \dfrac{x^3}{3} \\
&= \left(\dfrac{(1)^5}{5} - \dfrac{(1)^3}{3}\right) - \left(\dfrac{(-1)^5}{5} - \dfrac{(-1)^3}{3}\right) \\
% &= \dfrac{1}{5} - \dfrac{1}{3} + \dfrac{1}{5} - \dfrac{1}{3} \\
% &= \dfrac{3}{15} - \dfrac{5}{15} + \dfrac{3}{15} - \dfrac{5}{15} \\
&= -\dfrac{4}{15}; \quad \text{The area above the x-axis is less than the area below the x-axis} 
\end{align}
\end{problem}

\begin{problem}{39}
\begin{verbatim}

julia> import Pkg; Pkg.add("QuadGK"); using QuadGK
julia> f(x) = x - x^2
julia> quadgk(f, 0, 1.5)
(0.0, 0.0)
\end{verbatim}
\end{problem}

\begin{problem}{41}
\begin{verbatim}

julia> f(x) = x^4 - x^2
julia> quadgk(f, -1, 1)
(-0.26666666666666666, 0.0)
\end{verbatim}
\end{problem}

\begin{problem}{43}
$\displaystyle\int_1^3 (3t^2 + 7) \, dt$
\begin{align}
&= 3\dfrac{x^3}{3} + 7x \\
&= x^3 + 7x \\
&= ((3)^3 + 7(3)) - ((1)^3 + 7(1)) \\
&= 40
\end{align}
\end{problem}

\begin{problem}{45}
$\displaystyle\int_1^4 (\sqrt{x} - 1) \, dx$
\begin{align}
&= \dfrac{2}{3} x^{3/2} - x \\
&= \left(\dfrac{2}{3} (4)^{3/2} - (4)\right) - \left(\dfrac{2}{3} (1)^{3/2} - (1)\right) \\
&= \dfrac{5}{3}
\end{align}
\end{problem}

\begin{problem}{47}
$\displaystyle\int_{-2}^5 (2x^2 - 3x + 7) \, dx$
\begin{align}
&= \dfrac{2}{3} x^3 - \dfrac{3}{2} x^2 + 7x \\
&= \left(\dfrac{2}{3} (5)^3 - \dfrac{3}{2} (5)^2 + 7(5)\right) - \left(\dfrac{2}{3} (-2)^3 - \dfrac{3}{2} (-2)^2 + 7(-2)\right) \\
&= 106 \dfrac{1}{6}
\end{align}
\end{problem}

\begin{problem}{49}
$\displaystyle\int_{-5}^2 e^t \, dt$
\begin{align}
&= e^t \\
&= e^{(2)} - e^{(-5)} \\
&\approx 7.382
\end{align}
\end{problem}

\begin{problem}{51}
$\displaystyle\int_a^b \dfrac{1}{2} x^2 \, dx$
\begin{align}
&= \dfrac{1}{6} x^3 \, dx \\
&= \left(\dfrac{1}{6} (b)^3\right) - \left(\dfrac{1}{6} (a)^3\right) \\
&= \dfrac{b^3 - a^3}{6}
\end{align}
\end{problem}

\begin{problem}{53}
$\displaystyle\int_a^b e^{2t} \, dt$
\begin{align}
&= \dfrac{1}{2} e^{2t} \\
&= \left(\dfrac{1}{2} e^{2(b)}\right) - \left(\dfrac{1}{2} e^{2(a)}\right) \\
&= \dfrac{e^{2b} - e^{2a}}{2}
\end{align}
\end{problem}

\begin{problem}{55}
$\displaystyle\int_1^e \left(x + \dfrac{1}{x}\right) \, dx$
\begin{align}
&= \dfrac{x^2}{2} + \ln{x} \\
&= \left(\dfrac{(e)^2}{2} + \ln{2}\right) - \left(\dfrac{(1)^2}{2} + \ln{1}\right) \\
&\approx 4.195
\end{align}
\end{problem}

\begin{problem}{57}
$\displaystyle\int_0^2 \sqrt{2x} \, dx$
\begin{align}
&= \dfrac{2\sqrt{2} x^{3/2}}{3} \\
&= \left(\dfrac{2\sqrt{2} (2)^{3/2}}{3}\right) - \left(\dfrac{2\sqrt{2} (0)^{3/2}}{3}\right) \\
&= \dfrac{8}{3}
\end{align}
\end{problem}

\begin{problem}{73}
$v(t) = 3t^2, \quad s(0) = 4$
\begin{align}
s(t) &= \dfrac{3t^3}{3} + C \\
&= t^3 + C \\
4 &= (0)^3 + C \\
C &= 4 \\
s(t) &= t^3 + 4
\end{align}
\end{problem}

\begin{problem}{75}
$a(t) = 4t, \quad v(0) = 20$
\begin{align}
v(t) &= \dfrac{4t^2}{2} + C \\
&= 2t^2 + C \\
20 &= 2(0)^2 + C \\
C &= 20 \\
v(t) &= 2t^2 + 20
\end{align}
\end{problem}

\begin{problem}{93}
$\displaystyle\int_2^3 \dfrac{x^2 - 1}{x - 1} \, dx$
\begin{align}
&= \int_2^3 \dfrac{(x + 1)(x - 1)}{x - 1} \, dx \\
&= \int_2^3 (x - 1) \, dx \\
&= \dfrac{x^2}{2} - x \\
&= \left(\dfrac{(3)^2}{2} - (3)\right) - \left(\dfrac{(2)^2}{2} - (2)\right) \\
&= 1.5
\end{align}
\end{problem}

\begin{problem}{97}
$\displaystyle\int_1^8 \dfrac{\sqrt[3]{x^2} - 1}{\sqrt[3]{x}} \, dx$
\begin{align}
&= \int_1^8 (x^{2/3} - 1) \cdot x^{-1/3} \, dx \\
&= \int_1^8 x^{1/3} - x^{-1/3} \, dx \\
&= \dfrac{3x^{4/3}}{4} - \dfrac{3x^{2/3}}{2} \\
&= 6.75
\end{align}
\end{problem}

\begin{problem}{99}
$\displaystyle\int_2^5 (t + \sqrt{3})(t - \sqrt{3}) \, dt$
\begin{align}
&= \int_2^5 t^2 - 3 \, dt \\
&= \dfrac{t^3}{3} - 3t \\
&= \left(\dfrac{(5)^3}{3} - 3(5)\right) - \left(\dfrac{(2)^3}{3} - 3(2)\right) \\
&= 30
\end{align}
\end{problem}

\end{document}