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+% vim: set ts=2 sw=2 et tw=80:
+
+\documentclass[12pt,a4paper]{article}
+
+\usepackage[utf8]{inputenc} \usepackage[margin=2cm]{geometry}
+\usepackage{amstext} \usepackage{amsmath} \usepackage{array}
+\newcommand{\lra}{\Leftrightarrow}
+
+\title{Howework 5 -- Introduction to Computational Science}
+
+\author{Claudio Maggioni}
+
+\begin{document} \maketitle
+\section*{Question 1}
+Given the definition of degree of exactness being the highest polynomial degree
+$n$ at which a quadrature, for every polynomial of degree $n$, produces exactly
+the same polynomial, these are the proofs.
+
+\subsection*{Midpoint rule}
+All polynomials of degree 1 can be expressed as:
+
+\[p_1(x) = a_1 \cdot x + a_0\]
+
+Therefore their integral is:
+
+\[\int_0^1 a_1 \cdot x + a_0 dx = \frac{a_1}{2} + a_0\]
+
+The midpoint rule for $p_1(x)$ is
+
+\[f\left(\frac{1}{2}\right) \cdot 1 = \frac{a_1}{2} + a_0 = \int_0^1 a_1 \cdot x
++ a_0 dx\]
+
+Therefore the midpoint rule has a degree of exactness of at least 1.
+
+It is easy to show that the degree of exactness is not higher than 1 by
+considering the degree 2
+polynomial $x^2$, which has an integral in $[0, 1]$ of $\frac{1}{3}$ but a midpoint rule
+quadrature of $\frac{1}{4}$.
+
+\subsection*{Trapezoidal rule}
+The proof is similar to the one for the midpoint rule, but with this quadrature
+for degree 1 polynomials:
+
+\[\frac{f(0)}{2} + \frac{f(1)}{2} = \frac{a_0 + a_1 + a_0}{2} = \frac{a_1}{2} +
+a_0\]
+
+Which is again equal to the general integral for these polynomials.
+
+Again $x^2$ is a degree 2 polynomial with integral $\frac{1}{3}$ but a midpoint quadrature
+of $\frac{0 + 1}{2} = \frac{1}{2}$, thus bounding the degree of exactness to 1.
+
+\subsection*{Simpson rule}
+The proof is again similar, but for degree 3 polynomials which can all be
+written as:
+
+\[p_3(x) = a_3 x^3 + a_2 x^2 + a_1 x + a_0\]
+
+The integral is:
+
+\[\int_0^1p_3(x) dx = \frac{a_3}{4} + \frac{a_2}{3} + \frac{a_1}{2} + a_0\]
+
+The Simpson rule gives:
+
+\[\frac{1}{6} \cdot f(0) + \frac{4}{6}\cdot  f\left(\frac{1}{2}\right) + \frac{1}{6} \cdot f(1) =
+\frac{1}{6} a_0 + \frac{4}{6} \left(\frac{a_3}{8} + \frac{a_2}{4} +
+\frac{a_1}{2} + a_0\right) + \]\[\frac{1}{6} \left(a_3 + a_2 + a_1 + a_0\right) =
+\frac{a_3}{4} + \frac{a_2}{3} + \frac{a_1}{2} + a_0 = \int_0^1 p_3(x)dx\]
+
+Which tells us that the degree of exactness is at least 1.
+
+We can bound the degree of exactness to 3 with the 4th degree polynomial $x^4$
+which has integral in $[0, 1]$ of $\frac{1}{5}$ but has a quadrature of
+$\frac{1}{6} \cdot 0 + \frac{2}{3} \cdot \frac{1}{16} + \frac{1}{6} \cdot 1 =
+\frac{5}{24}$.
+\end{document}