Evaluate the following integrals. \int e^{3-4x}dx

Evaluate the following integrals.
$\int {e}^{3-4x}dx$
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Bubich13
Step 1
Given integral:
$\int {e}^{3-4x}dx$
Substitute t=3-4x and differentiate it w.r.t "x"
$\frac{dt}{dx}=\frac{d}{dx}\left(3-4x\right)$
$\frac{dt}{dx}=-4$
$\frac{-dt}{4}=dx$
Therefore, the given integral becomes,
$\frac{-1}{4}\int {e}^{t}dt$
Step 2
We know that
$\int {e}^{x}dx={e}^{x}+c$
Where, "c" is integration constant.
$⇒\frac{-1}{4}{e}^{t}+c$
Substitute t=3-4x in above equation, we get
$⇒\frac{-1}{4}{e}^{\left(3-4x\right)}+c$
$⇒\frac{-{e}^{\left(3-4x\right)}}{4}+c$
Not exactly what you’re looking for?
Bubich13
Given
$\int {e}^{3-4x}dx$
$\int -\frac{{e}^{t}}{4}dt$
$-\frac{1}{4}\cdot \int {e}^{t}dt$
Solve
$-\frac{1}{4}{e}^{t}$
$-\frac{1}{4}{e}^{3-4x}$
$-\frac{{e}^{3-4x}}{4}$
$-\frac{{e}^{3-4x}}{4}+C$
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Vasquez

$\begin{array}{}\int {e}^{3-4x}dx\\ =-\frac{1}{4}\int {e}^{u}du\\ \int {e}^{u}du\\ ={e}^{u}\\ -\frac{1}{4}\int {e}^{u}du\\ =-\frac{{e}^{u}}{4}\\ u=3-4x:\\ =-\frac{{e}^{3-4x}}{4}\\ Solution:\\ =-\frac{{e}^{3-4x}}{4}+C\end{array}$