Science:Math Exam Resources/Courses/MATH105/April 2016/Question 01 (l)/Solution 1

The region bounded by the function and x-axis from $[a,\infty )$ is $\int _{a}^{\infty }e^{-3x}dx.$

We first note that this is an improper integral, and we need to rewrite it as the following:

$\int _{a}^{\infty }e^{-3x}dx=\lim \limits _{b\to \infty }\int _{a}^{b}e^{-3x}dx.$

We now apply the substitution $u=3x$ to simplify the integral. Note that $du=3dx,$ and since $x$ is changing between $a$ and $b,$ then $u=3x$ is changing between $3a$ and $3b$.

Therefore,

$\int _{a}^{\infty }e^{-3x}dx=\lim \limits _{b\to \infty }\int _{a}^{b}e^{-3x}dx=\lim \limits _{b\to \infty }{\frac {1}{3}}\int _{3a}^{3b}e^{-u}du$ = $\lim \limits _{b\to \infty }{\frac {-1}{3}}e^{-u}|_{3a}^{3b}=\lim \limits _{b\to \infty }{\big [}{\frac {-1}{3}}e^{-3b}-{\frac {-1}{3}}e^{-3a}{\big ]}={\frac {1}{3}}e^{-3a}.$

We know that the area of the region is 1. This implies that

$1=\int _{a}^{\infty }e^{-3x}dx={\frac {1}{3}}e^{-3a}.$

We finally get

$a=-{\frac {1}{3}}ln3$