Notes: `y = 3x^2 - 2x, y = x^3 - 3x + 4` Use a graph to find approximate x-coordinates of the points of intersection of the given curves. Then find...

Sunday, 12 March 2017

$\displaystyle{y}={3}{{x}}^{{2}}-{2}{x},{y}={{x}}^{{3}}-{3}{x}+{4}$ Use a graph to find approximate x-coordinates of the points of intersection of the given curves. Then find...

$\displaystyle{y}={3}{{x}}^{{2}}-{2}{x},{y}={{x}}^{{3}}-{3}{x}+{4}$

Refer the attached image. Graph of $\displaystyle{y}={3}{{x}}^{{2}}-{2}{x}$ is plotted in red color and graph of $\displaystyle{y}={{x}}^{{3}}-{3}{x}+{4}$ is plotted in blue color.

From the graph, the x-coordinates of the intersection of curves are x $\displaystyle\approx$ -1.1 , x $\displaystyle\approx$ 1.25 , x $\displaystyle\approx$ 2.875.

Area of the region bounded by the curves A $\displaystyle={\int_{{-{1.1}}}^{{{1.25}}}}{\left({\left({{x}}^{{3}}-{3}{x}+{4}\right)}-{\left({3}{{x}}^{{2}}-{2}{x}\right)}\right)}{\left.{d}{x}\right.}+{\int_{{1.25}}^{{2.875}}}{\left({\left({3}{{x}}^{{2}}-{2}{x}\right)}-{\left({{x}}^{{3}}-{3}{x}+{4}\right)}\right)}{\left.{d}{x}\right.}$

$\displaystyle={\int_{{-{1.1}}}^{{1.25}}}{\left({{x}}^{{3}}-{3}{x}+{4}-{3}{{x}}^{{2}}+{2}{x}\right)}{\left.{d}{x}\right.}+{\int_{{1.25}}^{{2.875}}}{\left({3}{{x}}^{{2}}-{2}{x}-{{x}}^{{3}}+{3}{x}-{4}\right)}{\left.{d}{x}\right.}$

$\displaystyle={\int_{{-{1.1}}}^{{1.25}}}{\left({{x}}^{{3}}-{3}{{x}}^{{2}}-{x}+{4}\right)}{\left.{d}{x}\right.}+{\int_{{1.25}}^{{2.875}}}{\left(-{{x}}^{{3}}+{3}{{x}}^{{2}}+{x}-{4}\right)}{\left.{d}{x}\right.}$

$\displaystyle={{\left[\frac{{{x}}^{{4}}}{{4}}-{3}\frac{{{x}}^{{3}}}{{3}}-\frac{{{x}}^{{2}}}{{2}}+{4}{x}\right]}_{{-{1.1}}}^{{1.25}}}+{{\left[-\frac{{{x}}^{{4}}}{{4}}+{3}\frac{{{x}}^{{3}}}{{3}}+\frac{{{x}}^{{2}}}{{2}}-{4}{x}\right]}_{{1.25}}^{{2.875}}}$

$\displaystyle={{\left[\frac{{{x}}^{{4}}}{{4}}-{{x}}^{{3}}-\frac{{{x}}^{{2}}}{{2}}+{4}{x}\right]}_{{-{1.1}}}^{{1.25}}}+{{\left[-\frac{{{x}}^{{4}}}{{4}}+{{x}}^{{3}}+\frac{{{x}}^{{2}}}{{2}}-{4}{x}\right]}_{{1.25}}^{{2.875}}}$

$\displaystyle={\left(\frac{{{1.25}}^{{4}}}{{4}}-{{1.25}}^{{3}}-\frac{{{1.25}}^{{2}}}{{2}}+{4}\cdot{1.25}\right)}-{\left(\frac{{{\left(-{1.1}\right)}}^{{4}}}{{4}}-{{\left(-{1.1}\right)}}^{{3}}-\frac{{{\left(-{1.1}\right)}}^{{2}}}{{2}}+{4}{\left(-{1.1}\right)}\right)}+{\left(-\frac{{{\left({2.875}\right)}}^{{4}}}{{4}}+{{2.875}}^{{3}}+\frac{{{2.875}}^{{2}}}{{2}}-{4}{\left({2.875}\right)}\right)}-{\left(-\frac{{{\left({1.25}\right)}}^{{4}}}{{4}}+{{1.25}}^{{3}}+\frac{{{1.25}}^{{2}}}{{2}}-{4}{\left({1.25}\right)}\right)}$ $\displaystyle={\left({2.875976563}\right)}-{\left(-{3.307975}\right)}+{\left(-{0.683654785}\right)}-{\left(-{2.875976563}\right)}$ $\displaystyle={8.37627334}$

$\displaystyle{y}={3}{{x}}^{{2}}-{2}{x},{y}={{x}}^{{3}}-{3}{x}+{4}$

Refer the attached image. Graph of $\displaystyle{y}={3}{{x}}^{{2}}-{2}{x}$ is plotted in red color and graph of $\displaystyle{y}={{x}}^{{3}}-{3}{x}+{4}$ is plotted in blue color.

From the graph, the x-coordinates of the intersection of curves are x $\displaystyle\approx$ -1.1 , x $\displaystyle\approx$ 1.25 , x $\displaystyle\approx$ 2.875.