Question Number 1778 by 112358 last updated on 23/Sep/15 | ||
$${Let}\:{P}\left({t}\right)\:{denote}\:{a}\:{given}\:{cubic} \\ $$$${polynomial}.\:{Find}\:{the}\:{constants} \\ $$$${a}_{\mathrm{1}} ,{u}_{\mathrm{1}} ,{a}_{\mathrm{2}} \:{and}\:{u}_{\mathrm{2}} \:{such}\:{that} \\ $$$$\int_{−\mathrm{1}} ^{\:\mathrm{1}} {P}\left({t}\right){dt}={a}_{\mathrm{1}} {P}\left({u}_{\mathrm{1}} \right)+{a}_{\mathrm{2}} {P}\left({u}_{\mathrm{2}} \right). \\ $$ | ||
Answered by Rasheed Soomro last updated on 23/Sep/15 | ||
$${Let}\:{P}\left({t}\right)={at}^{\mathrm{3}} +{bt}^{\mathrm{2}} +{ct}+{d} \\ $$$${LHS}: \\ $$$$\:\:\:\:\:\:\int{P}\left({t}\right)\boldsymbol{\mathrm{d}}{t}=\frac{{a}}{\mathrm{4}}{t}^{\mathrm{4}} +\frac{{b}}{\mathrm{3}}{t}^{\mathrm{3}} +\frac{{c}}{\mathrm{2}}{t}^{\mathrm{2}} +{dt}+{C} \\ $$$$\:\:\:\:\:\:\:\int_{−\mathrm{1}} ^{\:\mathrm{1}} {P}\left({t}\right)\boldsymbol{\mathrm{d}}{t}=\left[\frac{{a}}{\mathrm{4}}{t}^{\mathrm{4}} +\frac{{b}}{\mathrm{3}}{t}^{\mathrm{3}} +\frac{{c}}{\mathrm{2}}{t}^{\mathrm{2}} +{dt}+{C}\right]_{−\mathrm{1}} ^{\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left[\frac{{a}}{\mathrm{4}}+\frac{{b}}{\mathrm{3}}+\frac{{c}}{\mathrm{2}}+{d}+{C}\right]−\left[\frac{{a}}{\mathrm{4}}−\frac{{b}}{\mathrm{3}}+\frac{{c}}{\mathrm{2}}−{d}+{C}\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{2}{b}}{\mathrm{3}}+\mathrm{2}{d} \\ $$$${RHS}: \\ $$$$\:\:\:\:\:\:\:{a}_{\mathrm{1}} {P}\left({u}_{\mathrm{1}} \right)+{a}_{\mathrm{2}} {P}\left({u}_{\mathrm{2}} \right) \\ $$$$\:\:\:\:\:={a}_{\mathrm{1}} \left\{{a}\left({u}_{\mathrm{1}} \right)^{\mathrm{3}} +{b}\left({u}_{\mathrm{1}} \right)^{\mathrm{2}} +{c}\left({u}_{\mathrm{1}} \right)+{d}\right\}+{a}_{\mathrm{2}} \left\{{a}\left({u}_{\mathrm{2}} \right)^{\mathrm{3}} +{b}\left({u}_{\mathrm{2}} \right)^{\mathrm{2}} +{cu}_{\mathrm{2}} +{d}\right\} \\ $$$${Comparing}\:{coefficients}\:{of}\:{b}\:\:{and}\:\:{d}\:{in}\:{LHS}\:\:{and}\:\:{RHS} \\ $$$$\:\:\:\frac{\mathrm{2}}{\mathrm{3}}={a}_{\mathrm{1}} \left({u}_{\mathrm{1}} \right)^{\mathrm{2}} +{a}_{\mathrm{2}} \left({u}_{\mathrm{2}} \right)^{\mathrm{2}} \:\:\:\:{and}\:\:\:\mathrm{2}={a}_{\mathrm{1}} +{a}_{\mathrm{2}} \\ $$$${Solving}\:{simultaneously} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{3}\left({u}_{\mathrm{1}} \right)^{\mathrm{2}} \:{a}_{\mathrm{1}} +\mathrm{3}\left({u}_{\mathrm{1}} \right)^{\mathrm{2}} {a}_{\mathrm{2}} =\mathrm{6}\left({u}_{\mathrm{1}} \right)^{\mathrm{2}} \:\:\:\:............{I} \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{3}\left({u}_{\mathrm{1}} \right)^{\mathrm{2}} {a}_{\mathrm{1}} +\mathrm{3}\left({u}_{\mathrm{2}} \right)^{\mathrm{2}} {a}_{\mathrm{2}} =\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:............{II} \\ $$$$\:{Subtracting}\:\:{II}\:\:\:{from}\:\:\:{I} \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{3}\left\{\left({u}_{\mathrm{1}} \right)^{\mathrm{2}} −\left({u}_{\mathrm{2}} \right)^{\mathrm{2}} \right\}{a}_{\mathrm{2}} =\mathrm{6}\left({u}_{\mathrm{1}} \right)^{\mathrm{2}} −\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{a}_{\mathrm{2}} =\frac{\mathrm{6}\left({u}_{\mathrm{1}} \right)^{\mathrm{2}} −\mathrm{2}}{\mathrm{3}\left\{\left({u}_{\mathrm{1}} \right)^{\mathrm{2}} −\left({u}_{\mathrm{2}} \right)^{\mathrm{2}} \right\}} \\ $$$${Similarly},\:\:\:\:{a}_{\mathrm{1}} =\frac{\mathrm{6}\left({u}_{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{2}}{\mathrm{3}\left\{\left({u}_{\mathrm{2}} \right)^{\mathrm{2}} −\left({u}_{\mathrm{1}} \right)^{\mathrm{2}} \right\}} \\ $$$${I}\:{think}\:{u}_{\mathrm{1}} \:{and}\:{u}_{\mathrm{2}} \:{should}\:{be}\:{taken}\:{as}\:{arbitrary}\:{constants}. \\ $$$${Or}\:{otherwise}\:{if}\:{a}_{\mathrm{1}} \:{and}\:\:{a}_{\mathrm{2}} \:\:{are}\:{taken}\:{as}\:\:{arbitrary}\:{then} \\ $$$${u}_{\mathrm{1}} \:\:{and}\:\:{u}_{\mathrm{2}} \:\:{can}\:{be}\:{determined}. \\ $$$$ \\ $$ | ||