\documentclass{article}
\usepackage{amsmath}
\usepackage{amsthm}
\usepackage{amssymb}
\usepackage{enumerate}
\begin{document}
\flushleft\underline{Homework 1 --- MATH 1586 Spring 2018} \\
Recall that for $n\neq -1$, 
$$\displaystyle\int x^n \mathrm{d}x = \dfrac{x^{n+1}}{n+1}+C.$$
Recall that 
$$\displaystyle\int \dfrac{1}{x} \mathrm{d}x=\ln(x)+C.$$
Recall that 
$$\displaystyle\int e^x \mathrm{d}x = e^x+C.$$
Recall that 
$$\displaystyle\int f(x) + g(x) \mathrm{d}x=\displaystyle\int f(x) \mathrm{d}x + \displaystyle\int g(x) \mathrm{d}x.$$
Finally recall that if $F$ represents any antiderivative of $f$, then 
$$\displaystyle\int_a^b f(x) \mathrm{d}x = F(b)-F(a).$$
\noindent Compute the following definite and indefinite integrals.
\begin{enumerate}[1.]
\item $\displaystyle\int x^{17} \mathrm{d}x$ \\
\textit{Solution}: Calculate
$$\displaystyle\int x^{17} \mathrm{d}x = \dfrac{x^{18}}{18}+C.$$
\item $\displaystyle\int x^3+3x^2+\dfrac{x}{9}+1 \mathrm{d}x$ \\
\textit{Solution}: Calculate 
$$\displaystyle\int x^3+3x^2+\dfrac{x}{9}+1 \mathrm{d}x = \dfrac{x^4}{4} + x^3 + \dfrac{x^2}{18} + x + C.$$
\item $\displaystyle\int \dfrac{e^x}{10} + x^2 + \dfrac{9}{x} + 1 \mathrm{d}x$ \\
\textit{Solution}: Calculate
$$\begin{array}{ll}
\displaystyle\int \dfrac{e^x}{10} + x^2 + \dfrac{9}{x} + 1 \mathrm{d}x &+ \dfrac{1}{10} \displaystyle\int e^x \mathrm{d}x + \displaystyle\int x^2 \mathrm{d}x + 9 \displaystyle\int \dfrac{1}{x} \mathrm{d}x + \displaystyle\int 1 \mathrm{d}x \\
&= \dfrac{e^x}{10} + \dfrac{x^3}{3} + 9\ln(x)+ x + C.
\end{array}$$
\item $\displaystyle\int_{0}^1 4x^4+3x^2+1 \mathrm{d}x$ \\
\textit{Solution}: Calculate
$$\begin{array}{ll}
\displaystyle\int_0^1 4x^4 + 3x^2 + 1 \mathrm{d}x &= \dfrac{4x^5}{5} + x^3 + x \Bigg|_0^1 \\
&= \left( \dfrac{4}{5} + 1^3 + 1 \right) - \left( \dfrac{4(0^5)}{5} + 0^3 + 0 \right) \\
&= \dfrac{14}{5}.
\end{array}$$
\item $\displaystyle\int_{-3}^4 -3e^x + \dfrac{1}{x} + x^5 - 2x^2 \mathrm{d}x$ \\
\textit{Solution}: Calculate
\begin{eqnarray*}
\lefteqn{\displaystyle\int_{-3}^4 -3e^x + \dfrac{1}{x} + x^5 - 2x^2 \mathrm{d}x} \\
& &= -3e^x + \ln(x)+\dfrac{x^6}{6} - \dfrac{2x^3}{3} \Bigg|_{-3}^4 \\
& &= \left( -3e^4 + \ln(4) + \dfrac{4^6}{6} - \dfrac{2(4^3)}{3} \right) - \left( -3 e^{-3} + \ln(-3) + \dfrac{(-3)^6}{6} - \dfrac{2(-3)^3}{3} \right).
\end{eqnarray*}
(\textit{note: ``$\ln(-3)$" is a nebulous concept -- technically we should not be able to integrate this as I did above because of the singularity at $x=0$ -- oops! If the lower and upper bounds were positive, the technique would work fine.})
\item Is it accurate to say that 
$$(*) \hspace{35pt} \displaystyle\int e^{7x} \mathrm{d}x=e^{7x}+C?$$
How can you use differentiation to check if this is an accurate statement? \\
\textit{Solution}:  No! It is easy to see that $e^{7x}+C$ is not the antiderivative of $e^{7x}$. It can be checked by differentiating $e^{7x}+C$:
$$\dfrac{\mathrm{d}}{\mathrm{d}x} \Bigg[ e^{7x}+C \Bigg] = 7e^{7x}.$$
This shows that the formula $(*)$ is not correct, because the right-hand side would have to equal $e^{7x}$ (i.e. it would have to be the function inside the integral). 
\end{enumerate}
\end{document}