# Segment 4: The Jailer's Tip - 1/25/2013

Problem 1
First, we need to make our integral into the form of $\int_{-\infty}^\infty \delta(u)du$
$\int_0^1 \delta(3x-2) dx$
Let u = 3x - 2
du = 3dx
dx = du/3

To change our bounds into terms of u:

• For 0: 3(0) - 2 = -2
• For 1: 3(1) - 2 = 1

We know that $\int_{-\infty}^\infty \delta(u)du = 1$ since only at 0 we get 1. Thus, our integral becomes
$\frac{1}{3}\cdot \int_{-2}^1 \delta(u) du = 1 \cdot \frac{1}{3} = \frac{1}{3}$

Problem 2: Prove that $\delta(ax) = \frac{1}{a}\delta(x)$

$\int_{-\infty}^\infty \delta(ax)du$
Let u = ax
du = a dx
dx = du/a
$\frac{1}{a}\int_{-\infty}^\infty \delta(u)du$ = $\bold{\frac{1}{a}}$

$\int_{-\infty}^\infty \frac{1}{a}\delta(x) = \frac{1}{a}\int_{-\infty}^\infty \delta(x)$ = $\bold{\frac{1}{a}}$

Problem 3:

• $\int_0^1 \frac{1}{1 + x } \cdot \frac{1}{2} \cdot (\delta(x - \frac{1}{3} )+ \delta(x - \frac{2}{3})dx)$
• $\frac{1}{2}\cdot (\frac{1}{1 + \frac{1}{3}} + \frac{1}{1 + \frac{2}{3}})$
• $\frac{1}{2}\cdot (\frac{1}{\frac{4}{3}} + \frac{1}{\frac{5}{3}})$
• $\frac{1}{2}\cdot (\frac{3}{4} + \frac{3}{5})$
• $\frac{1}{2}\cdot (\frac{15}{20} + \frac{12}{20})$
• $\frac{1}{2}\cdot (\frac{27}{20})$
• $\frac{27}{40} = 0.675$