Frank Guyton

2021-12-31

Evaluate the definite integral.
${\int }_{-4}^{4}{3}^{\frac{x}{4}}dx$

John Koga

The given integral is ${\int }_{-4}^{4}{3}^{\frac{x}{4}}dx$
Substituting $\frac{x}{4}$ as t.
Therefore $\frac{dx}{4}=dt$
Evaluating the integral:
${\int }_{-4}^{4}{3}^{\frac{x}{4}}dx=4{\int }_{-1}^{1}{3}^{t}dt$
$=4{\left[\frac{{3}^{t}}{\mathrm{ln}3}\right]}_{-1}^{1}$
$=4\left[\frac{3}{\mathrm{ln}3}-\frac{{3}^{-1}}{\mathrm{ln}3}\right]$
$=\frac{4}{\mathrm{ln}3}\left(3-\frac{1}{3}\right)$
$=\frac{32}{3\mathrm{ln}3}$

kaluitagf

${\int }_{-4}^{4}{3}^{\frac{x}{4}}dx$
$\int {3}^{\frac{x}{4}}dx$
$\int {3}^{t}\cdot 4dt$
$4\cdot \int {3}^{t}dt$
$4\cdot \frac{{3}^{t}}{\mathrm{ln}\left(3\right)}$
$4\cdot \frac{{3}^{\frac{x}{4}}}{\mathrm{ln}\left(3\right)}$
Calculate
$\frac{4\cdot {3}^{\frac{x}{4}}}{\mathrm{ln}\left(3\right)}$
Return the limits
$\frac{4\cdot {3}^{\frac{x}{4}}}{\mathrm{ln}\left(3\right)}{\mid }_{-4}^{4}$
Calculate the expression
$\frac{4\cdot {3}^{\frac{4}{4}}}{\mathrm{ln}\left(3\right)}-\frac{4\cdot {3}^{\frac{-4}{4}}}{\mathrm{ln}\left(3\right)}$
Simplify
$\frac{32}{3\mathrm{ln}\left(3\right)}$

Vasquez

${\int }_{-4}^{4}{3}^{x/4}dx$
This is a tabular integral:
$\int {3}^{x/4}dx=\frac{4\ast {3}^{x/4}}{\mathrm{ln}\left(3\right)}+C$
Let's calculate a definite integral:
$\begin{array}{}Given:\\ {\int }_{-4}^{4}{3}^{x/4}dx=\left(\frac{4\ast {3}^{x/4}}{\mathrm{ln}\left(3\right)}\right){|}_{-4}^{4}\\ F\left(4\right)=\frac{12}{\mathrm{ln}\left(3\right)}\\ F\left(-4\right)=\frac{4}{3\ast \mathrm{ln}\left(3\right)}\\ I=\frac{12}{\mathrm{ln}\left(3\right)}0\left(\frac{4}{3\ast \mathrm{ln}\left(3\right)}\right)=\frac{32}{3\ast \mathrm{ln}\left(3\right)}\end{array}$

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