Integration Questions

Question Number 53078 by gunawan last updated on 16/Jan/19

$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\left({x}^{\mathrm{3}} +\mathrm{1}\right)^{\mathrm{3}/\mathrm{2}} }\:{dx}=... \\$$

Commented byMJS last updated on 17/Jan/19

$$\approx.\mathrm{774606} \\$$

Answered by tanmay.chaudhury50@gmail.com last updated on 17/Jan/19

$$\mathrm{1}+{x}^{\mathrm{3}} >\mathrm{1}+{x}^{\mathrm{2}} \\$$ $$\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{3}} }<\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} } \\$$ $${but}\:{in}\:{the}\:{interval}\:\left[\mathrm{0},\mathrm{1}\right] \\$$ $$\frac{\mathrm{1}}{\left(\mathrm{1}+{x}^{\mathrm{3}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} }>\frac{\mathrm{1}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} {so}} \\$$ $${i}\:{am}\:{rectifying} \\$$ $$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{3}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} }>\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} } \\$$ $${now}\: \\$$ $$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} } \\$$ $${x}={tana}\:\:\: \\$$ $$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{{sec}^{\mathrm{2}} {a}}{\left(\mathrm{1}+{tan}^{\mathrm{2}} {a}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{da} \\$$ $$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{{sec}^{\mathrm{2}} {ada}}{{sec}^{\mathrm{3}} {a}} \\$$ $$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {cosada} \\$$ $$\mid{sina}\mid_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \\$$ $$=\frac{\mathrm{1}}{\sqrt{\mathrm{2}}} \\$$ $${so}\:\:\:\:\:\:{I}>\frac{\mathrm{1}}{\sqrt{\mathrm{2}}} \\$$ $${I}>\mathrm{0}.\mathrm{71} \\$$ $${attaching}\:{graph}... \\$$ $${answer}\:{yet}\:{to}\:{find}.... \\$$ $$\\$$ $$\\$$

Commented bytanmay.chaudhury50@gmail.com last updated on 17/Jan/19

Commented bytanmay.chaudhury50@gmail.com last updated on 17/Jan/19

$${area}\:{of}\:{trapazium}\:{calculation} \\$$ $${whenf}\left({x}\right)=\frac{\mathrm{1}}{\left(\mathrm{1}+{x}^{\mathrm{3}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} } \\$$ $${at}\:{x}=\mathrm{1}\:\:\:{f}\left(\mathrm{1}\right)=\frac{\mathrm{1}}{\left(\mathrm{2}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}} \\$$ $${at}\:{x}=\mathrm{0}\:\:{f}\left(\mathrm{0}\right)=\frac{\mathrm{1}}{\left(\mathrm{1}+\mathrm{0}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }=\mathrm{1} \\$$ $${so}\:{area}\:{of}\:{trapazium}=\frac{\mathrm{1}}{\mathrm{2}}\left[{f}\left(\mathrm{0}\right)+{f}\left(\mathrm{1}\right)\right]×\mathrm{1} \\$$ $$=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\right)×\mathrm{1}=\mathrm{0}.\mathrm{68} \\$$ $${from}\:{graph}\:{it}\:{is}\:{clear}\:{that}\:{area}\:{bounded}\:{i},{e} \\$$ $$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{3}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} }>\mathrm{0}.\mathrm{68} \\$$ $$\\$$