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+++ b/hw5/hw5.tex
@@ -72,4 +72,41 @@ 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
 $\frac{1}{6} \cdot 0 + \frac{2}{3} \cdot \frac{1}{16} + \frac{1}{6} \cdot 1 =
 \frac{5}{24}$.
+
+\section*{Question 2}
+The algebraic solution is:
+
+\[\int_0^1 1 - 4(x - 0.5)^2 dx = 1 - 4 \cdot \int_0^1 x^2 - x + \frac14 =
+1 - 4 \left(\frac{4-6+3}{12}\right) = 1 - \frac13 = \frac23\]
+
+The solution using quadrature is:
+
+\[Q = \frac{1}{2} (f(0) + f(1)) = \frac{1}{2}(0 + 0) = 0 \qquad (x, h) =
+\left(\frac{1}{2}, \frac{1}{2}\right) \qquad \epsilon = \frac{1}{10}\]
+\[E\left(\frac{1}{2}, \frac12\right) = f\left(\frac12\right) - \frac12\left(f(0)
++ f(1)\right) = 1 > \frac1{10} \qquad Q = 0 + \frac12 \cdot 1 = \frac12\]
+
+\[E\left(\frac{1}{4}, \frac14\right) = f\left(\frac14\right) - \frac12\left(f(0)
++ f\left(\frac12\right)\right) = \frac34 - \frac12 = \frac14 > \frac1{10} \qquad
+Q = \frac12 + \frac14 \cdot \frac14 = \frac9{16}\]
+
+\[E\left(\frac{3}{4}, \frac14\right) = f\left(\frac34\right) -
+\frac12\left(f\left(\frac12\right)
++ f(1)\right) = \frac14 > \frac1{10} \qquad
+Q = \frac9{16} + \frac14 \cdot \frac14 = \frac{10}{16}\]
+
+\[E\left(\frac18, \frac18\right) = f\left(\frac18\right) -
+\frac12\left(f\left(0\right)
++ f\left(\frac14\right)\right) = \frac7{16} - \frac12 \cdot \frac34 = \frac1{16} < \frac1{10}\]
+\[E\left(\frac38, \frac18\right) = f\left(\frac38\right) -
+\frac12\left(f\left(\frac14\right)
++ f\left(\frac12\right)\right) = \frac{15}{16} - \frac12 \cdot \frac74 = \frac1{16} < \frac1{10}\]
+\[E\left(\frac58, \frac18\right) = f\left(\frac58\right) -
+\frac12\left(f\left(\frac12\right)
++ f\left(\frac34\right)\right) = \frac{15}{16} - \frac12 \cdot \frac74 = \frac1{16} < \frac1{10}\]
+\[E\left(\frac78, \frac18\right) = f\left(\frac78\right) -
+\frac12\left(f\left(\frac34\right)
++ f\left(1\right)\right) = \frac7{16} - \frac12 \cdot \frac34 = \frac1{16} < \frac1{10}\]
+
+Thus the solution using quadrature is $\frac{5}{8}$.
 \end{document}