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Question-89781




Question Number 89781 by mr W last updated on 19/Apr/20
Commented by jagoll last updated on 19/Apr/20
A= ((Δ(√Δ))/(6.a^2 ))
$$\mathrm{A}=\:\frac{\Delta\sqrt{\Delta}}{\mathrm{6}.\mathrm{a}^{\mathrm{2}} } \\ $$
Commented by mr W last updated on 29/Apr/20
how to quickly calculate the area   bounded by two parabolas or by a  parabola and a straight line    recently a lot of such questions are  asked.  the general way to solve such questions  is to apply intergral calculus. but in  case of parabola, we can find some  short cuts. see also Q88758.    since a straight line is also a special  parabola, we will only look at the case  with two parabolas.    parabola 1: y_1 =a_1 x^2 +b_1 x+c_1   parabola 2: y_2 =a_2 x^2 +b_2 x+c_2   Δy=y_1 −y_2 =(a_1 −a_2 )x^2 +(b_1 −b_2 )x+(c_1 −c_2 )  or  Δy=ax^2 +bx+c   with a=a_1 −a_2 , b=b_1 −b_2 , c=c_1 −c_2   we see Δy represents also a parabola.    the area between both parabolas is:  (we use ∣ ∣ since the integral can also  be negative, only its absolute value is  the area)  A=∣∫_x_P  ^x_Q  (y_1 −y_2 )dx∣  A=∣∫_x_P  ^x_Q  (Δy)dx∣  A=∣∫_x_P  ^x_Q  (ax^2 +bx+c)dx∣  A=∣(a/3)(x_Q ^3 −x_P ^3 )+(b/2)(x_Q ^2 −x_P ^2 )+c(x_Q −x_P )∣  A=∣{(a/3)(x_Q ^2 +x_P x_Q +x_P ^2 )+(b/2)(x_Q +x_P )+c}(x_Q −x_P )∣  A=∣{(a/3)[(x_P +x_Q )^2 −x_P x_Q ]+(b/2)(x_Q +x_P )+c}(x_Q −x_P )∣    since x_P  and x_Q  are the roots of eqn.  ax^2 +bx+c=0  we have  D=b^2 −4ac (usually we use Δ instead of D,  here i use D because i have used Δ above)  x_P +x_Q =−(b/a)  x_P x_Q =(c/a)  x_Q −x_P =(√((x_Q +x_P )^2 −4x_P x_Q ))=(√((b^2 /a^2 )−((4c)/a)))=((√(b^2 −4ac))/a)=((√δ)/a)  A=∣{(a/3)[(−(b/a))^2 −(c/a)]+(b/2)(−(b/a))+c}((√D)/a)∣  A=∣(−b^2 +4ac)((√D)/(6a^2 ))∣  A=(b^2 −4ac)((√D)/(6a^2 ))  ⇒A=((D(√D))/(6a^2 )) with D=b^2 −4ac
$${how}\:{to}\:{quickly}\:{calculate}\:{the}\:{area}\: \\ $$$${bounded}\:{by}\:{two}\:{parabolas}\:{or}\:{by}\:{a} \\ $$$${parabola}\:{and}\:{a}\:{straight}\:{line} \\ $$$$ \\ $$$${recently}\:{a}\:{lot}\:{of}\:{such}\:{questions}\:{are} \\ $$$${asked}. \\ $$$${the}\:{general}\:{way}\:{to}\:{solve}\:{such}\:{questions} \\ $$$${is}\:{to}\:{apply}\:{intergral}\:{calculus}.\:{but}\:{in} \\ $$$${case}\:{of}\:{parabola},\:{we}\:{can}\:{find}\:{some} \\ $$$${short}\:{cuts}.\:{see}\:{also}\:{Q}\mathrm{88758}. \\ $$$$ \\ $$$${since}\:{a}\:{straight}\:{line}\:{is}\:{also}\:{a}\:{special} \\ $$$${parabola},\:{we}\:{will}\:{only}\:{look}\:{at}\:{the}\:{case} \\ $$$${with}\:{two}\:{parabolas}. \\ $$$$ \\ $$$${parabola}\:\mathrm{1}:\:{y}_{\mathrm{1}} ={a}_{\mathrm{1}} {x}^{\mathrm{2}} +{b}_{\mathrm{1}} {x}+{c}_{\mathrm{1}} \\ $$$${parabola}\:\mathrm{2}:\:{y}_{\mathrm{2}} ={a}_{\mathrm{2}} {x}^{\mathrm{2}} +{b}_{\mathrm{2}} {x}+{c}_{\mathrm{2}} \\ $$$$\Delta{y}={y}_{\mathrm{1}} −{y}_{\mathrm{2}} =\left({a}_{\mathrm{1}} −{a}_{\mathrm{2}} \right){x}^{\mathrm{2}} +\left({b}_{\mathrm{1}} −{b}_{\mathrm{2}} \right){x}+\left({c}_{\mathrm{1}} −{c}_{\mathrm{2}} \right) \\ $$$${or} \\ $$$$\Delta{y}={ax}^{\mathrm{2}} +{bx}+{c}\: \\ $$$${with}\:{a}={a}_{\mathrm{1}} −{a}_{\mathrm{2}} ,\:{b}={b}_{\mathrm{1}} −{b}_{\mathrm{2}} ,\:{c}={c}_{\mathrm{1}} −{c}_{\mathrm{2}} \\ $$$${we}\:{see}\:\Delta{y}\:{represents}\:{also}\:{a}\:{parabola}. \\ $$$$ \\ $$$${the}\:{area}\:{between}\:{both}\:{parabolas}\:{is}: \\ $$$$\left({we}\:{use}\:\mid\:\mid\:{since}\:{the}\:{integral}\:{can}\:{also}\right. \\ $$$${be}\:{negative},\:{only}\:{its}\:{absolute}\:{value}\:{is} \\ $$$$\left.{the}\:{area}\right) \\ $$$${A}=\mid\int_{{x}_{{P}} } ^{{x}_{{Q}} } \left({y}_{\mathrm{1}} −{y}_{\mathrm{2}} \right){dx}\mid \\ $$$${A}=\mid\int_{{x}_{{P}} } ^{{x}_{{Q}} } \left(\Delta{y}\right){dx}\mid \\ $$$${A}=\mid\int_{{x}_{{P}} } ^{{x}_{{Q}} } \left({ax}^{\mathrm{2}} +{bx}+{c}\right){dx}\mid \\ $$$${A}=\mid\frac{{a}}{\mathrm{3}}\left({x}_{{Q}} ^{\mathrm{3}} −{x}_{{P}} ^{\mathrm{3}} \right)+\frac{{b}}{\mathrm{2}}\left({x}_{{Q}} ^{\mathrm{2}} −{x}_{{P}} ^{\mathrm{2}} \right)+{c}\left({x}_{{Q}} −{x}_{{P}} \right)\mid \\ $$$${A}=\mid\left\{\frac{{a}}{\mathrm{3}}\left({x}_{{Q}} ^{\mathrm{2}} +{x}_{{P}} {x}_{{Q}} +{x}_{{P}} ^{\mathrm{2}} \right)+\frac{{b}}{\mathrm{2}}\left({x}_{{Q}} +{x}_{{P}} \right)+{c}\right\}\left({x}_{{Q}} −{x}_{{P}} \right)\mid \\ $$$${A}=\mid\left\{\frac{{a}}{\mathrm{3}}\left[\left({x}_{{P}} +{x}_{{Q}} \right)^{\mathrm{2}} −{x}_{{P}} {x}_{{Q}} \right]+\frac{{b}}{\mathrm{2}}\left({x}_{{Q}} +{x}_{{P}} \right)+{c}\right\}\left({x}_{{Q}} −{x}_{{P}} \right)\mid \\ $$$$ \\ $$$${since}\:{x}_{{P}} \:{and}\:{x}_{{Q}} \:{are}\:{the}\:{roots}\:{of}\:{eqn}. \\ $$$${ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0} \\ $$$${we}\:{have} \\ $$$${D}={b}^{\mathrm{2}} −\mathrm{4}{ac}\:\left({usually}\:{we}\:{use}\:\Delta\:{instead}\:{of}\:{D},\right. \\ $$$$\left.{here}\:{i}\:{use}\:{D}\:{because}\:{i}\:{have}\:{used}\:\Delta\:{above}\right) \\ $$$${x}_{{P}} +{x}_{{Q}} =−\frac{{b}}{{a}} \\ $$$${x}_{{P}} {x}_{{Q}} =\frac{{c}}{{a}} \\ $$$${x}_{{Q}} −{x}_{{P}} =\sqrt{\left({x}_{{Q}} +{x}_{{P}} \right)^{\mathrm{2}} −\mathrm{4}{x}_{{P}} {x}_{{Q}} }=\sqrt{\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }−\frac{\mathrm{4}{c}}{{a}}}=\frac{\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}}{{a}}=\frac{\sqrt{\delta}}{{a}} \\ $$$${A}=\mid\left\{\frac{{a}}{\mathrm{3}}\left[\left(−\frac{{b}}{{a}}\right)^{\mathrm{2}} −\frac{{c}}{{a}}\right]+\frac{{b}}{\mathrm{2}}\left(−\frac{{b}}{{a}}\right)+{c}\right\}\frac{\sqrt{{D}}}{{a}}\mid \\ $$$${A}=\mid\left(−{b}^{\mathrm{2}} +\mathrm{4}{ac}\right)\frac{\sqrt{{D}}}{\mathrm{6}{a}^{\mathrm{2}} }\mid \\ $$$${A}=\left({b}^{\mathrm{2}} −\mathrm{4}{ac}\right)\frac{\sqrt{{D}}}{\mathrm{6}{a}^{\mathrm{2}} } \\ $$$$\Rightarrow{A}=\frac{{D}\sqrt{{D}}}{\mathrm{6}{a}^{\mathrm{2}} }\:{with}\:{D}={b}^{\mathrm{2}} −\mathrm{4}{ac} \\ $$
Commented by mr W last updated on 19/Apr/20
i know jagoll sir uses this formula.   here i just wanted to show how the  formula comes. i think people apply  better, if they understand. that′s why  i not only want to share the formula,  but also to tell why.
$${i}\:{know}\:{jagoll}\:{sir}\:{uses}\:{this}\:{formula}.\: \\ $$$${here}\:{i}\:{just}\:{wanted}\:{to}\:{show}\:{how}\:{the} \\ $$$${formula}\:{comes}.\:{i}\:{think}\:{people}\:{apply} \\ $$$${better},\:{if}\:{they}\:{understand}.\:{that}'{s}\:{why} \\ $$$${i}\:{not}\:{only}\:{want}\:{to}\:{share}\:{the}\:{formula}, \\ $$$${but}\:{also}\:{to}\:{tell}\:{why}. \\ $$
Commented by jagoll last updated on 19/Apr/20
yes sir thank you
$$\mathrm{yes}\:\mathrm{sir}\:\mathrm{thank}\:\mathrm{you} \\ $$
Commented by I want to learn more last updated on 19/Apr/20
It really helped me sir. More knowledge sir.  I really appreciate sir.
$$\mathrm{It}\:\mathrm{really}\:\mathrm{helped}\:\mathrm{me}\:\mathrm{sir}.\:\mathrm{More}\:\mathrm{knowledge}\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}\:\mathrm{sir}. \\ $$
Commented by M±th+et£s last updated on 19/Apr/20
this is verry usefull for me god bless  you sir.
$${this}\:{is}\:{verry}\:{usefull}\:{for}\:{me}\:{god}\:{bless} \\ $$$${you}\:{sir}. \\ $$
Commented by otchereabdullai@gmail.com last updated on 19/Apr/20
wonderful
$$\mathrm{wonderful} \\ $$
Commented by Coronavirus last updated on 27/Jun/20
thanks so much sir

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