New Notes Blog: `int_0^2(y - 1)(2y + 1)dy` Evaluate the integral.

Saturday, 11 April 2015

$\displaystyle{\int_{{0}}^{{2}}}{\left({y}-{1}\right)}{\left({2}{y}+{1}\right)}{\left.{d}{y}\right.}$ Evaluate the integral.

$\displaystyle{\int_{{0}}^{{2}}}{\left({y}-{1}\right)}{\left({2}{y}+{1}\right)}{\left.{d}{y}\right.}$

Before evaluating, expand the integrand.

$\displaystyle={\int_{{0}}^{{2}}}{\left({2}{{y}}^{{2}}+{y}-{2}{y}-{1}\right)}{\left.{d}{y}\right.}$

$\displaystyle={\int_{{0}}^{{2}}}{\left({2}{{y}}^{{2}}-{y}-{1}\right)}{\left.{d}{y}\right.}$

Then, apply the integral formulas $\displaystyle\int{{x}}^{{n}}{\left.{d}{x}\right.}=\frac{{{x}}^{{{n}+{1}}}}{{{n}+{1}}}$ and $\displaystyle\int{c}{\left.{d}{x}\right.}={c}{x}$ .

$\displaystyle={\left(\frac{{{2}{{y}}^{{3}}}}{{3}}-\frac{{{y}}^{{2}}}{{2}}-{y}\right)}{{\mid}_{{0}}^{{2}}}$

Then, plug-in the limits of the integral to the resulting function.

$\displaystyle={\left(\frac{{{2}{{\left({2}\right)}}^{{3}}}}{{3}}-\frac{{{2}}^{{2}}}{{2}}-{2}\right)}-{\left(\frac{{{2}{{\left({0}\right)}}^{{3}}}}{{3}}-\frac{{{0}}^{{2}}}{{2}}-{0}\right)}$

$\displaystyle={\left(\frac{{16}}{{3}}-\frac{{4}}{{2}}-{2}\right)}-{0}$

$\displaystyle=\frac{{16}}{{3}}-{2}-{2}$

$\displaystyle=\frac{{16}}{{3}}-{4}$

$\displaystyle=\frac{{16}}{{3}}-\frac{{12}}{{3}}$

$\displaystyle=\frac{{4}}{{3}}$

Therefore, $\displaystyle{\int_{{0}}^{{2}}}{\left({y}-{1}\right)}{\left({2}{y}+{1}\right)}{\left.{d}{y}\right.}=\frac{{4}}{{3}}$ .

$\displaystyle{\int_{{0}}^{{2}}}{\left({y}-{1}\right)}{\left({2}{y}+{1}\right)}{\left.{d}{y}\right.}$

Before evaluating, expand the integrand.

$\displaystyle={\int_{{0}}^{{2}}}{\left({2}{{y}}^{{2}}+{y}-{2}{y}-{1}\right)}{\left.{d}{y}\right.}$

$\displaystyle={\int_{{0}}^{{2}}}{\left({2}{{y}}^{{2}}-{y}-{1}\right)}{\left.{d}{y}\right.}$

Then, apply the integral formulas $\displaystyle\int{{x}}^{{n}}{\left.{d}{x}\right.}=\frac{{{x}}^{{{n}+{1}}}}{{{n}+{1}}}$ and $\displaystyle\int{c}{\left.{d}{x}\right.}={c}{x}$ .

$\displaystyle={\left(\frac{{{2}{{y}}^{{3}}}}{{3}}-\frac{{{y}}^{{2}}}{{2}}-{y}\right)}{{\mid}_{{0}}^{{2}}}$

Then, plug-in the limits of the integral to the resulting function.

$\displaystyle={\left(\frac{{{2}{{\left({2}\right)}}^{{3}}}}{{3}}-\frac{{{2}}^{{2}}}{{2}}-{2}\right)}-{\left(\frac{{{2}{{\left({0}\right)}}^{{3}}}}{{3}}-\frac{{{0}}^{{2}}}{{2}}-{0}\right)}$

$\displaystyle={\left(\frac{{16}}{{3}}-\frac{{4}}{{2}}-{2}\right)}-{0}$

$\displaystyle=\frac{{16}}{{3}}-{2}-{2}$

$\displaystyle=\frac{{16}}{{3}}-{4}$

$\displaystyle=\frac{{16}}{{3}}-\frac{{12}}{{3}}$

$\displaystyle=\frac{{4}}{{3}}$

Therefore, $\displaystyle{\int_{{0}}^{{2}}}{\left({y}-{1}\right)}{\left({2}{y}+{1}\right)}{\left.{d}{y}\right.}=\frac{{4}}{{3}}$ .

New Notes Blog

Saturday, 11 April 2015

$\displaystyle{\int_{{0}}^{{2}}}{\left({y}-{1}\right)}{\left({2}{y}+{1}\right)}{\left.{d}{y}\right.}$ Evaluate the integral.

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In "By the Waters of Babylon," under the leadership of John, what do you think the Hill People will do with their society?

Saturday, 11 April 2015

Evaluate the integral.

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Post a Comment

In &quot;By the Waters of Babylon,&quot; under the leadership of John, what do you think the Hill People will do with their society?

$\displaystyle{\int_{{0}}^{{2}}}{\left({y}-{1}\right)}{\left({2}{y}+{1}\right)}{\left.{d}{y}\right.}$ Evaluate the integral.

In "By the Waters of Babylon," under the leadership of John, what do you think the Hill People will do with their society?