# Using the substitution w = x^2, find a function g(w) such that \int_a^b e ^{-x} dx =...

## Question:

Using the substitution {eq}w = x^2 {/eq}, find a function {eq}g(w) {/eq} such that {eq}\int_a^b e ^{-x} dx = \int_{e^a}^{e^b} g(w) dx {/eq} for all {eq}a<b {/eq}.

## Integral by Substitution:

Let us consider the integral {eq}I = \int {f(g(x))} g'(x)dx {/eq}, where {eq}f(x) {/eq} and {eq}g(x) {/eq} are real valued functions of {eq}x {/eq}.

Let us assume the substitution {eq}z = g(x) \Rightarrow dz = g'(x)dx {/eq}

After substitution:

{eq}I = \int {f(z)} dz {/eq}, which is easily solvable.

## Answer and Explanation:

Here, in this case, the given integral can be represented as:

{eq}\eqalign{ \int_a^b {{e^{ - x}}} dx& = \int_{z = {e^a}}^{{e^b}} {{e^{ - \ln z}}} dx\,\,\,\,\left[ {{\text{Assume: }}x = \ln \left( z \right) \Rightarrow z = {e^x}} \right] \cr & = \int_{z = {e^a}}^{{e^b}} {\frac{1}{z}} \frac{1}{z}dz \cr & = \int_{z = {e^a}}^{{e^b}} {\frac{1}{{{z^2}}}} dz \cr & = \int_{x = {e^a}}^{{e^b}} {\frac{1}{{{x^2}}}} dx\, \cr & = \int_{x = {e^a}}^{{e^b}} {\frac{1}{w}} dx\,\,\,\left[ {{\text{Substitute: }}w = {x^2}} \right] \cr & \color{blue}{\therefore g\left( w \right) = \frac{1}{w}} \cr} {/eq}