Notes: `int_0^4 x/sqrt(3+2x) dx` Use integration tables to evaluate the definite integral.

Thursday, 22 January 2015

$\displaystyle{\int_{{0}}^{{4}}}\frac{{x}}{\sqrt{{{3}+{2}{x}}}}{\left.{d}{x}\right.}$ Use integration tables to evaluate the definite integral.

To evaluate the given integral problem: $\displaystyle{\int_{{0}}^{{4}}}\frac{{x}}{\sqrt{{{3}+{2}{x}}}}{\left.{d}{x}\right.}$ , we determine first the indefinite integral function F(x). From the table of indefinite integrals, we may consider the formula for integrals with roots as:

$\displaystyle\int\frac{{u}}{\sqrt{{{u}\pm{a}}}}{d}{u}=\frac{{2}}{{3}}{\left({u}-+{2}{a}\right)}\sqrt{{{u}\pm{a}}}+{C}$

Take note that we have " $\displaystyle+$ " sign inside the square root on $\displaystyle{\int_{{0}}^{{4}}}\frac{{x}}{\sqrt{{{3}+{2}{x}}}}{\left.{d}{x}\right.}$ then we will follow:

$\displaystyle\int\frac{{u}}{\sqrt{{{u}+{a}}}}{d}{u}=\frac{{2}}{{3}}{\left({u}-{2}{a}\right)}\sqrt{{{u}+{a}}}+{C}.$

We may let $\displaystyle{a}={3}$ and $\displaystyle{u}={2}{x}$ or $\displaystyle{x}=\ldots$

$\displaystyle\int\frac{{u}}{\sqrt{{{u}\pm{a}}}}{d}{u}=\frac{{2}}{{3}}{\left({u}-+{2}{a}\right)}\sqrt{{{u}\pm{a}}}+{C}$

Take note that we have " $\displaystyle+$ " sign inside the square root on $\displaystyle{\int_{{0}}^{{4}}}\frac{{x}}{\sqrt{{{3}+{2}{x}}}}{\left.{d}{x}\right.}$ then we will follow:

$\displaystyle\int\frac{{u}}{\sqrt{{{u}+{a}}}}{d}{u}=\frac{{2}}{{3}}{\left({u}-{2}{a}\right)}\sqrt{{{u}+{a}}}+{C}.$

We may let $\displaystyle{a}={3}$ and $\displaystyle{u}={2}{x}$ or $\displaystyle{x}=\frac{{u}}{{2}}$

For the derivative of u, we get $\displaystyle{d}{u}={2}{\left.{d}{x}\right.}$ or $\displaystyle\frac{{{d}{u}}}{{2}}={\left.{d}{x}\right.}$ .

Plug-in the values: $\displaystyle{u}={2}{x}$ or $\displaystyle{x}=\frac{{u}}{{2}}$ ,and $\displaystyle\frac{{{d}{u}}}{{2}}={\left.{d}{x}\right.}$ , we get:

$\displaystyle{\int_{{0}}^{{4}}}\frac{{x}}{\sqrt{{{3}+{2}{x}}}}{\left.{d}{x}\right.}={\int_{{0}}^{{4}}}\frac{{\frac{{u}}{{2}}}}{\sqrt{{{3}+{u}}}}\cdot\frac{{{d}{u}}}{{2}}$

$\displaystyle={\int_{{0}}^{{4}}}\frac{{{u}{d}{u}}}{{{4}\sqrt{{{3}+{u}}}}}$

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

$\displaystyle{\int_{{0}}^{{4}}}\frac{{{u}{d}{u}}}{{{4}\sqrt{{{3}+{u}}}}}=\frac{{1}}{{4}}{\int_{{0}}^{{4}}}\frac{{{u}{d}{u}}}{\sqrt{{{3}+{u}}}}$

Apply the aforementioned integral formula from the table of integrals, we get:

$\displaystyle\frac{{1}}{{4}}{\int_{{0}}^{{4}}}\frac{{{u}{d}{u}}}{\sqrt{{{3}+{u}}}}=\frac{{1}}{{4}}\cdot{\left[\frac{{2}}{{3}}{\left({u}-{2}{\left({3}\right)}\right)}\sqrt{{{u}+{3}}}\right]}{{\mid}_{{0}}^{{4}}}$

$\displaystyle=\frac{{1}}{{4}}\cdot{\left[\frac{{2}}{{3}}{\left({u}-{6}\right)}\sqrt{{{u}+{3}}}\right]}{{\mid}_{{0}}^{{4}}}$

$\displaystyle=\frac{{2}}{{12}}{\left({u}-{6}\right)}\sqrt{{{u}+{3}}}{{\mid}_{{0}}^{{4}}}$

$\displaystyle=\frac{{1}}{{6}}{\left({u}-{6}\right)}\sqrt{{{u}+{3}}}{\]}{{\left|_{{0}}^{{4}}{\quad\text{or}\quad}\frac{{{\left({u}-{6}\right)}\sqrt{{{u}+{3}}}}}{{6}}\right|}_{{0}}^{{4}}}$

Plug-in $\displaystyle{u}={2}{x}$ on $\displaystyle\frac{{{\left({u}-{6}\right)}\sqrt{{{u}+{3}}}}}{{6}}+{C}$ , we get:

$\displaystyle{\int_{{0}}^{{4}}}\frac{{x}}{\sqrt{{{3}+{2}{x}}}}{\left.{d}{x}\right.}=\frac{{{\left({2}{x}-{6}\right)}\sqrt{{{2}{x}+{3}}}}}{{6}}{{\mid}_{{0}}^{{4}}}$

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

$\displaystyle\frac{{{\left({2}{x}-{6}\right)}\sqrt{{{2}{x}+{3}}}}}{{6}}{{\mid}_{{0}}^{{4}}}=\frac{{{\left({2}{\left({4}\right)}-{6}\right)}\sqrt{{{2}{\left({4}\right)}+{3}}}}}{{6}}-\frac{{{\left({2}{\left({0}\right)}-{6}\right)}\sqrt{{{2}{\left({0}\right)}+{3}}}}}{{6}}$

$\displaystyle=\frac{{{\left({8}-{6}\right)}\sqrt{{{8}+{3}}}}}{{6}}-\frac{{{\left({0}-{6}\right)}\sqrt{{{0}+{3}}}}}{{6}}$

$\displaystyle=\frac{{{2}\cdot\sqrt{{{11}}}}}{{6}}-\frac{{-{6}\sqrt{{{3}}}}}{{6}}$

$\displaystyle=\frac{\sqrt{{{11}}}}{{3}}-{\left(-\sqrt{{{3}}}\right)}$

$\displaystyle=\frac{\sqrt{{{11}}}}{{3}}+\sqrt{{{3}}}$

$\displaystyle=\frac{{\sqrt{{{11}}}+{3}\sqrt{{{3}}}}}{{3}}$ or $\displaystyle{2.84}$ (approximated value).

Notes

Thursday, 22 January 2015

$\displaystyle{\int_{{0}}^{{4}}}\frac{{x}}{\sqrt{{{3}+{2}{x}}}}{\left.{d}{x}\right.}$ 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?

Thursday, 22 January 2015

Use integration tables to evaluate the definite integral.

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Post a Comment

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

$\displaystyle{\int_{{0}}^{{4}}}\frac{{x}}{\sqrt{{{3}+{2}{x}}}}{\left.{d}{x}\right.}$ Use integration tables to evaluate the definite integral.

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