Notes: `int_-1^2(1-x)dx` Evaluate the integral by interpreting it in terms of areas.

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

To
interpret the integral in terms of area , graph the integrand.

The integrand
is the function $\displaystyle{f{{\left({x}\right)}}}={1}-{x}$

Graph the function in the interval (-1,2). Refer
the attached graph.

The bounded region forms two triangles, one triangle
below the x-axis and second triangle above the x-axis.

Area of triangle above
the x-axis $\displaystyle{A}_{{2}}=\frac{{1}}{{2}}{b}_{{2}}{h}_{{2}}$

$\displaystyle{A}_{{2}}=\frac{{1}}{{2}}\cdot{2}\cdot{2}={2}$

Area of
triangle below the x-axis $\displaystyle{A}_{{1}}=\frac{{1}}{{2}}{b}_{{1}}{h}_{{1}}$

$\displaystyle{A}_{{1}}=\frac{{1}}{{2}}\cdot{1}\cdot{1}=\frac{{1}}{{2}}$

So, $\displaystyle{\int_{{-{{1}}}}^{{2}}}{\left({1}-{x}\right)}{\left.{d}{x}\right.}={A}_{{2}}-{A}_{{1}}$

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

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

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Notes

Thursday, 8 March 2018

$\displaystyle{\int_{{-{{1}}}}^{{2}}}{\left({1}-{x}\right)}{\left.{d}{x}\right.}$ Evaluate the integral by interpreting it in terms of areas.

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