Notes: `int_0^1 xe^(x^2) dx` Use integration tables to evaluate the definite integral.

Tuesday, 25 February 2014

$\displaystyle{\int_{{0}}^{{1}}}{x}{{e}}^{{{{x}}^{{2}}}}{\left.{d}{x}\right.}$ Use integration tables to evaluate the definite integral.

For the given problem: $\displaystyle{\int_{{0}}^{{1}}}{x}{{e}}^{{{{x}}^{{2}}}}$ , we may first solve for its indefinite integral. Indefinite integral are written in the form of $\displaystyle\int{f{{\left({x}\right)}}}{\left.{d}{x}\right.}={F}{\left({x}\right)}+{C}$

where: $\displaystyle{f{{\left({x}\right)}}}$ as the integrand

$\displaystyle{F}{\left({x}\right)}$ as the anti-derivative function

$\displaystyle{C}$ as the arbitrary constant known as constant of integration

We omit the arbitrary constant C when we have a boundary values: a to b. We...

where: $\displaystyle{f{{\left({x}\right)}}}$ as the integrand

$\displaystyle{F}{\left({x}\right)}$ as the anti-derivative function

$\displaystyle{C}$ as the arbitrary constant known as constant of integration

We omit the arbitrary constant C when we have a boundary values: a to b. We follow formula: $\displaystyle{\int_{{a}}^{{b}}}{f{{\left({x}\right)}}}{\left.{d}{x}\right.}={F}{\left({x}\right)}{{\mid}_{{a}}^{{b}}}$ .

Form the table of integrals, we follow the indefinite integral formula for exponential function as:

$\displaystyle\int{x}{{e}}^{{-{a}{{x}}^{{2}}}}{\left.{d}{x}\right.}=-\frac{{1}}{{{2}{a}}}{{e}}^{{-{a}{{x}}^{{2}}}}+{C}$

By comparison of $\displaystyle-{a}{{x}}^{{2}}$ with $\displaystyle{{x}}^{{2}}$ shows that we let $\displaystyle{a}=-{1}$ .

Plug-in $\displaystyle{a}=-{1}$ on $\displaystyle-{a}{{x}}^{{2}}$ for checking, we get: $\displaystyle-{\left(-{1}\right)}{{x}}^{{2}}=+{{x}}^{{2}}$ or $\displaystyle{{x}}^{{2}}$ .

Plug-in $\displaystyle{a}=-{1}$ on integral formula, we get:

$\displaystyle{\int_{{0}}^{{1}}}{x}{{e}}^{{{{x}}^{{2}}}}=-\frac{{1}}{{{2}{\left(-{1}\right)}}}{{e}}^{{{\left(-{\left(-{1}\right)}{{x}}^{{2}}\right)}}}{{\mid}_{{0}}^{{1}}}$

$\displaystyle=-\frac{{1}}{{-{2}}}{{e}}^{{{\left({1}\cdot{{x}}^{{2}}\right)}}}{{\mid}_{{0}}^{{1}}}$

$\displaystyle=\frac{{1}}{{2}}{{e}}^{{{{x}}^{{2}}}}{{\mid}_{{0}}^{{1}}}$

Applying definite integral formula: $\displaystyle{F}{\left({x}\right)}{{\mid}_{{a}}^{{b}}}={F}{\left({b}\right)}\equiv{F}{\left({a}\right)}$ .

$\displaystyle\frac{{1}}{{2}}{{e}}^{{{{x}}^{{2}}}}{{\mid}_{{0}}^{{1}}}=\frac{{1}}{{2}}{{e}}^{{{{1}}^{{2}}}}-\frac{{1}}{{2}}{{e}}^{{{{0}}^{{2}}}}$

$\displaystyle=\frac{{1}}{{2}}{{e}}^{{{1}}}-\frac{{1}}{{2}}{{e}}^{{{0}}}$

$\displaystyle=\frac{{1}}{{2}}{e}-\frac{{1}}{{2}}\cdot{1}$

$\displaystyle=\frac{{1}}{{2}}{e}-\frac{{1}}{{2}}{\quad\text{or}\quad}\frac{{1}}{{2}}{\left({e}-{1}\right)}$

Notes

Tuesday, 25 February 2014

$\displaystyle{\int_{{0}}^{{1}}}{x}{{e}}^{{{{x}}^{{2}}}}{\left.{d}{x}\right.}$ Use integration tables to evaluate the definite integral.

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How are race, gender, and class addressed in Oliver Optic's Rich and Humble?

Tuesday, 25 February 2014

Use integration tables to evaluate the definite integral.

No comments:

Post a Comment

How are race, gender, and class addressed in Oliver Optic&#39;s Rich and Humble?

$\displaystyle{\int_{{0}}^{{1}}}{x}{{e}}^{{{{x}}^{{2}}}}{\left.{d}{x}\right.}$ Use integration tables to evaluate the definite integral.

How are race, gender, and class addressed in Oliver Optic's Rich and Humble?