`int_-1^2 2^x dx` Evaluate the definite integral

Saturday, September 5, 2009

$\displaystyle{\int_{{-{{1}}}}^{{2}}}{{2}}^{{x}}{\left.{d}{x}\right.}$ Evaluate the definite integral

Recall the First Fundamental Theorem of Calculus:

If f is continuous on closed interval [a,b], we follow:

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

where F is the anti-derivative of f on [a,b].

This shows that we need to solve first the indefinite integral F(x) to be able to apply the difference of values F based on the given boundary limit of a and b.

The resulting value will be the definite integral.

For the given problem $\displaystyle{\int_{{-{1}}}^{{{2}}}}{{2}}^{{x}}{\left.{d}{x}\right.}$ , the integrand function $\displaystyle{f{{\left({x}\right)}}}={{2}}^{{x}}$

which is in a form of a exponential function.

The basic integration formula for exponential function follows:

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

By comparison: $\displaystyle{{a}}^{{u}}$ vs $\displaystyle{{2}}^{{x}}$ , we may let:

$\displaystyle{a}={2}$ , $\displaystyle{u}={x}$ and then $\displaystyle{d}{u}={\left.{d}{x}\right.}$

Then applying the formula, we get:

$\displaystyle\int{{2}}^{{x}}{\left.{d}{x}\right.}=\frac{{{2}}^{{x}}}{{{\ln{{\left({2}\right)}}}}}$

indefinite integral function $\displaystyle{F}{\left({x}\right)}=\frac{{{2}}^{{x}}}{{{\ln{{\left({2}\right)}}}}}$

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

Based on the given problem: $\displaystyle{\int_{{-{1}}}^{{{2}}}}{{2}}^{{x}}{\left.{d}{x}\right.}$ , the boundary limits are:

lower limit: $\displaystyle{a}=-{1}$ and upper limit: $\displaystyle{b}={2}$

Plug-in the boundary limits in $\displaystyle{F}{\left({x}\right)}=\frac{{{2}}^{{x}}}{{{\ln{{\left({2}\right)}}}}}$ one at a time, we get:

$\displaystyle{F}{\left({a}\right)}={F}{\left(-{1}\right)}=\frac{{{{2}}^{{-{1}}}}}{{\ln{{\left({2}\right)}}}}$

$\displaystyle{F}{\left({a}\right)}{F}{\left(-{1}\right)}=\frac{{1}}{{{2}{\ln{{\left({2}\right)}}}}}$

$\displaystyle{F}{\left({b}\right)}={F}{\left({2}\right)}=\frac{{{2}}^{{2}}}{{{\ln{{\left({2}\right)}}}}}$

$\displaystyle{F}{\left({b}\right)}=\frac{{4}}{{{\ln{{\left({2}\right)}}}}}$

Solving for the definite integral:

$\displaystyle{F}{\left({b}\right)}-{F}{\left({a}\right)}={F}{\left({2}\right)}-{F}{\left(-{1}\right)}$

$\displaystyle=\frac{{4}}{{{\ln{{\left({2}\right)}}}}}-\frac{{1}}{{{2}{\ln{{\left({2}\right)}}}}}$

$\displaystyle={4}\cdot\frac{{1}}{{{\ln{{\left({2}\right)}}}}}-{\left(\frac{{1}}{{2}}\right)}\cdot\frac{{1}}{{{\ln{{\left({2}\right)}}}}}$

$\displaystyle={\left({4}-\frac{{1}}{{2}}\right)}\cdot\frac{{1}}{{{\ln{{\left({2}\right)}}}}}$

$\displaystyle=\frac{{7}}{{2}}\cdot\frac{{1}}{{{\ln{{\left({2}\right)}}}}}$

or $\displaystyle\frac{{7}}{{{2}{\ln{{\left({2}\right)}}}}}$ as the Final Answer.

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Saturday, September 5, 2009

$\displaystyle{\int_{{-{{1}}}}^{{2}}}{{2}}^{{x}}{\left.{d}{x}\right.}$ Evaluate the definite integral

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

Saturday, September 5, 2009

Evaluate the definite integral

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

$\displaystyle{\int_{{-{{1}}}}^{{2}}}{{2}}^{{x}}{\left.{d}{x}\right.}$ 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?