New Notes Blog: `int_-1^2(x^3 - 2x)dx` Evaluate the integral.

Friday, 28 March 2014

$\displaystyle{\int_{{-{{1}}}}^{{2}}}{\left({{x}}^{{3}}-{2}{x}\right)}{\left.{d}{x}\right.}$ Evaluate the integral.

You need to evaluate the integral, such that:

$\displaystyle{\int_{{-{1}}}^{{2}}}{\left({{x}}^{{3}}-{2}{x}\right)}{\left.{d}{x}\right.}={\int_{{-{1}}}^{{2}}}{{x}}^{{3}}{\left.{d}{x}\right.}-{\int_{{-{1}}}^{{2}}}{2}{x}{\left.{d}{x}\right.}$

$\displaystyle{\int_{{-{1}}}^{{2}}}{\left({{x}}^{{3}}-{2}{x}\right)}{\left.{d}{x}\right.}={\left(\frac{{{x}}^{{4}}}{{4}}-{{x}}^{{2}}\right)}{{\mid}_{{-{1}}}^{{2}}}$

$\displaystyle{\int_{{-{1}}}^{{2}}}{\left({{x}}^{{3}}-{2}{x}\right)}{\left.{d}{x}\right.}={\left(\frac{{{2}}^{{4}}}{{4}}-{2}\cdot{2}-\frac{{{\left(-{1}\right)}}^{{4}}}{{4}}-{2}\right)}$

$\displaystyle{\int_{{-{1}}}^{{2}}}{\left({{x}}^{{3}}-{2}{x}\right)}{\left.{d}{x}\right.}={4}-{4}-\frac{{1}}{{4}}-{2}$

$\displaystyle{\int_{{-{1}}}^{{2}}}{\left({{x}}^{{3}}-{2}{x}\right)}{\left.{d}{x}\right.}=-\frac{{9}}{{4}}$

Hence evauating the definite integral yields $\displaystyle{\int_{{-{1}}}^{{2}}}{\left({{x}}^{{3}}-{2}{x}\right)}{\left.{d}{x}\right.}=-\frac{{9}}{{4}}.$

You need to evaluate the integral, such that:

$\displaystyle{\int_{{-{1}}}^{{2}}}{\left({{x}}^{{3}}-{2}{x}\right)}{\left.{d}{x}\right.}={\int_{{-{1}}}^{{2}}}{{x}}^{{3}}{\left.{d}{x}\right.}-{\int_{{-{1}}}^{{2}}}{2}{x}{\left.{d}{x}\right.}$

$\displaystyle{\int_{{-{1}}}^{{2}}}{\left({{x}}^{{3}}-{2}{x}\right)}{\left.{d}{x}\right.}={\left(\frac{{{x}}^{{4}}}{{4}}-{{x}}^{{2}}\right)}{{\mid}_{{-{1}}}^{{2}}}$

$\displaystyle{\int_{{-{1}}}^{{2}}}{\left({{x}}^{{3}}-{2}{x}\right)}{\left.{d}{x}\right.}={\left(\frac{{{2}}^{{4}}}{{4}}-{2}\cdot{2}-\frac{{{\left(-{1}\right)}}^{{4}}}{{4}}-{2}\right)}$

$\displaystyle{\int_{{-{1}}}^{{2}}}{\left({{x}}^{{3}}-{2}{x}\right)}{\left.{d}{x}\right.}={4}-{4}-\frac{{1}}{{4}}-{2}$

$\displaystyle{\int_{{-{1}}}^{{2}}}{\left({{x}}^{{3}}-{2}{x}\right)}{\left.{d}{x}\right.}=-\frac{{9}}{{4}}$

Hence evauating the definite integral yields $\displaystyle{\int_{{-{1}}}^{{2}}}{\left({{x}}^{{3}}-{2}{x}\right)}{\left.{d}{x}\right.}=-\frac{{9}}{{4}}.$

New Notes Blog

Friday, 28 March 2014

$\displaystyle{\int_{{-{{1}}}}^{{2}}}{\left({{x}}^{{3}}-{2}{x}\right)}{\left.{d}{x}\right.}$ 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?

Friday, 28 March 2014

Evaluate the integral.

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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?

$\displaystyle{\int_{{-{{1}}}}^{{2}}}{\left({{x}}^{{3}}-{2}{x}\right)}{\left.{d}{x}\right.}$ 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?