Question Number 11315 by b.e.h.i.8.3.4.1.7@gmail.com last updated on 20/Mar/17
Commented by chux last updated on 20/Mar/17
$$\mathrm{please}\:\mathrm{can}\:\mathrm{you}\:\mathrm{tell}\:\mathrm{me}\:\mathrm{the}\:\mathrm{name}\:\mathrm{of} \\ $$$$\mathrm{any}\:\mathrm{app}\:\mathrm{for}\:\mathrm{plotting}\:\mathrm{and}\:\mathrm{editing} \\ $$$$\mathrm{graph}. \\ $$
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 20/Mar/17
Commented by mrW1 last updated on 20/Mar/17
$${B}\left(−\mathrm{1},−\mathrm{1}\right) \\ $$$${A}\left(\mathrm{1},\mathrm{1}\right) \\ $$$${y}={x}^{\mathrm{3}} \\ $$$${y}'=\mathrm{3}{x}^{\mathrm{2}} \\ $$$${L}=\int_{−\mathrm{1}} ^{\mathrm{1}} \sqrt{\mathrm{1}+\left({y}'\right)^{\mathrm{2}} }{dx} \\ $$$${L}=\int_{−\mathrm{1}} ^{\mathrm{1}} \sqrt{\mathrm{1}+\mathrm{9}{x}^{\mathrm{4}} }{dx} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\int\sqrt{\mathrm{1}+\left(\sqrt{\mathrm{3}}{x}\right)^{\mathrm{4}} }{d}\left(\sqrt{\mathrm{3}}{x}\right) \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\int\sqrt{\mathrm{1}+{t}^{\mathrm{4}} }{dt}\:\:\:{with}\:{t}=\sqrt{\mathrm{3}}{x} \\ $$$${I}=\int\sqrt{\mathrm{1}+{t}^{\mathrm{4}} }{dt} \\ $$$$ \\ $$$${working}… \\ $$
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 20/Mar/17
$${ok}.{thanks}.{but}\:{i}\:{have}\:{problem}\:{with} \\ $$$${this}\:{integral}! \\ $$
Commented by Joel576 last updated on 20/Mar/17
$$\mathrm{GeoGebra} \\ $$
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 21/Mar/17
$${hello}\:{dear}.{i}\:{can}\:{not}\:{solve}\:{this}\:{integral} \\ $$$${please}\:{help}\:{me}. \\ $$