`int_-4^4 3^(x/4) dx` Evaluate the definite integral

Wednesday, June 27, 2012

$\displaystyle{\int_{{-{{4}}}}^{{4}}}{{3}}^{{\frac{{x}}{{4}}}}{\left.{d}{x}\right.}$ Evaluate the definite integral

To evaluate the integral: $\displaystyle{\int_{{-{4}}}^{{{4}}}}{{3}}^{{\frac{{x}}{{4}}}}{\left.{d}{x}\right.}$ , we follow the formula based from the First Fundamental Theorem of Calculus:

$\displaystyle{\int_{{a}}^{{b}}}{f{{\left({x}\right)}}}{\left.{d}{x}\right.}={F}{\left({b}\right)}-{F}{\left({a}\right)}$

wherein f is a continuous and F is the indefinite integral f on the closed interval [a,b].

Based on the given problem, the boundary limits are:

a =-4 and b=4

To solve for F as the indefinite integral of f, we follow the basic integration formula for an exponential function:

$\displaystyle\int{{a}}^{{u}}{d}{u}=\frac{{{a}}^{{u}}}{{{\ln{{\left({a}\right)}}}}}+{C}$

By comparison: $\displaystyle{{a}}^{{u}}$ vs $\displaystyle{{3}}^{{\frac{{x}}{{4}}}}$ , we let:

$\displaystyle{a}={3}$ and $\displaystyle{u}=\frac{{x}}{{4}}$ then $\displaystyle{d}{u}=\frac{{1}}{{4}}{\left.{d}{x}\right.}$ .

Rearrange $\displaystyle{d}{u}=\frac{{1}}{{4}}{\left.{d}{x}\right.}$ into $\displaystyle{4}{d}{u}={\left.{d}{x}\right.}$ .

Apply u-substitution using $\displaystyle{u}=\frac{{x}}{{4}}$ and $\displaystyle{4}{d}{u}={\left.{d}{x}\right.}$ :

$\displaystyle\int{{3}}^{{\frac{{x}}{{4}}}}{\left.{d}{x}\right.}=\int{{3}}^{{u}}\cdot{4}{d}{u}$

Apply the basic properties of integration: $\displaystyle\int{c}\cdot{f{{\left({x}\right)}}}{\left.{d}{x}\right.}={c}\int{f{{\left({x}\right)}}}{\left.{d}{x}\right.}$ .

$\displaystyle\int{{3}}^{{u}}\cdot{4}{d}{u}={4}\int{{3}}^{{u}}{d}{u}$

Applying the formula: $\displaystyle\int{{a}}^{{u}}{d}{u}=\frac{{{a}}^{{u}}}{{{\ln{{\left({a}\right)}}}}}$ .

$\displaystyle{4}\int{{3}}^{{u}}{d}{u}={4}\cdot{\left[\frac{{{3}}^{{u}}}{{{\ln{{\left({3}\right)}}}}}\right]}$

Express in terms of x using $\displaystyle{u}=\frac{{x}}{{4}}$ :

$\displaystyle{4}\cdot{\left[\frac{{{3}}^{{u}}}{{{\ln{{\left({3}\right)}}}}}\right]}={4}\cdot{\left[\frac{{{3}}^{{\frac{{x}}{{4}}}}}{{{\ln{{\left({3}\right)}}}}}\right]}$

Then indefinite integral function $\displaystyle{F}{\left({x}\right)}={4}\cdot{\left[\frac{{{3}}^{{\frac{{x}}{{4}}}}}{{{\ln{{\left({3}\right)}}}}}\right]}$

Applying F(b) - F(a) with the closed interval [a,b] as [-4,4]:

$\displaystyle{\int_{{-{4}}}^{{{4}}}}{{3}}^{{\frac{{x}}{{4}}}}{\left.{d}{x}\right.}={4}\cdot{\left[\frac{{{3}}^{{\frac{{{4}}}{{4}}}}}{{{\ln{{\left({3}\right)}}}}}\right]}-{4}\cdot{\left[\frac{{{3}}^{{-\frac{{4}}{{4}}}}}{{{\ln{{\left({3}\right)}}}}}\right]}$

$\displaystyle=\frac{{{4}\cdot{{3}}^{{{1}}}}}{{{\ln{{\left({3}\right)}}}}}-\frac{{{4}\cdot{{3}}^{{-{1}}}}}{{{\ln{{\left({3}\right)}}}}}$

$\displaystyle=\frac{{{12}}}{{{\ln{{\left({3}\right)}}}}}-\frac{{4}}{{{3}{\ln{{\left({3}\right)}}}}}$

$\displaystyle={\left({12}-\frac{{4}}{{3}}\right)}\cdot\frac{{1}}{{{\ln{{\left({3}\right)}}}}}$

$\displaystyle={\left(\frac{{{36}}}{{3}}-\frac{{4}}{{3}}\right)}\cdot\frac{{1}}{{{\ln{{\left({3}\right)}}}}}$

$\displaystyle={\left(\frac{{32}}{{3}}\right)}\cdot\frac{{1}}{{{\ln{{\left({3}\right)}}}}}$ or $\displaystyle\frac{{{32}}}{{{3}{\ln{{\left({3}\right)}}}}}$ as the Final Answer.

Blog

Wednesday, June 27, 2012

$\displaystyle{\int_{{-{{4}}}}^{{4}}}{{3}}^{{\frac{{x}}{{4}}}}{\left.{d}{x}\right.}$ Evaluate the definite integral

No comments:

Post a Comment

Thomas Jefferson's election in 1800 is sometimes called the Revolution of 1800. Why could it be described in this way?

Wednesday, June 27, 2012

Evaluate the definite integral

No comments:

Post a Comment

Thomas Jefferson&#39;s election in 1800 is sometimes called the Revolution of 1800. Why could it be described in this way?

$\displaystyle{\int_{{-{{4}}}}^{{4}}}{{3}}^{{\frac{{x}}{{4}}}}{\left.{d}{x}\right.}$ Evaluate the definite integral

Thomas Jefferson's election in 1800 is sometimes called the Revolution of 1800. Why could it be described in this way?