Question Number 203881 by mr W last updated on 31/Jan/24
Commented by mr W last updated on 31/Jan/24
$${find}\:{the}\:{hatched}\:{area}\:{between}\:{curve} \\ $$$${and}\:{its}\:{tangent}\:{line}. \\ $$
Answered by Frix last updated on 01/Feb/24
$$\mathrm{Curve}:\:{f}\left({x}\right)={x}^{\mathrm{4}} +{ax}^{\mathrm{3}} +{bx}^{\mathrm{2}} +{cx}+{d} \\ $$$$\mathrm{Tangent}:\:{g}\left({x}\right)={px}+{q} \\ $$$${f}\left(\mathrm{3}\right)={g}\left(\mathrm{3}\right) \\ $$$${f}\left(\mathrm{9}\right)={g}\left(\mathrm{9}\right) \\ $$$${f}'\left(\mathrm{3}\right)={g}'\left(\mathrm{3}\right) \\ $$$${f}'\left(\mathrm{9}\right)={g}'\left(\mathrm{9}\right) \\ $$$$\Rightarrow \\ $$$${f}\left({x}\right)={x}^{\mathrm{4}} −\mathrm{24}{x}^{\mathrm{3}} +\mathrm{198}{x}^{\mathrm{2}} +{cx}+{d} \\ $$$${g}\left({x}\right)=\mathrm{648}{x}−\mathrm{729}+{cx}+{d} \\ $$$$\underset{\mathrm{3}} {\overset{\mathrm{9}} {\int}}{f}\left({x}\right)−{g}\left({x}\right){dx}=\frac{\mathrm{1296}}{\mathrm{5}} \\ $$
Commented by mr W last updated on 01/Feb/24
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Answered by mr W last updated on 01/Feb/24
Commented by mr W last updated on 01/Feb/24
![curve: f(x)=x^4 +ax^3 +bx^2 +cx+d tangent: g(x)=hx+k f(x)−g(x) is also a quatric equation which has two double real roots: x=p=3, x=q=9 ⇒f(x)−g(x)=(x−p)^2 (x−q)^2 A=∫_p ^q [f(x)−g(x)]dx =∫_p ^q (x−p)^2 (x−q)^2 dx =∫_0 ^(q−p) u^2 (u−(q−p))^2 du with u=x−q =∫_0 ^s u^2 (u−s)^2 du with s=q−p =∫_0 ^s (u^4 −2su^3 +s^2 u^2 )du =(s^5 /5)−(s^5 /2)+(s^5 /3) =(s^5 /(30)) with p=3, q=9 ⇒s=9−3=6 ⇒A=(6^5 /(30))=((1296)/5) ✓](https://www.tinkutara.com/question/Q203894.png)
$${curve}:\:{f}\left({x}\right)={x}^{\mathrm{4}} +{ax}^{\mathrm{3}} +{bx}^{\mathrm{2}} +{cx}+{d} \\ $$$${tangent}:\:{g}\left({x}\right)={hx}+{k} \\ $$$${f}\left({x}\right)−{g}\left({x}\right)\:{is}\:{also}\:{a}\:{quatric}\:{equation} \\ $$$${which}\:{has}\:{two}\:{double}\:{real}\:{roots}:\: \\ $$$${x}={p}=\mathrm{3},\:{x}={q}=\mathrm{9} \\ $$$$\Rightarrow{f}\left({x}\right)−{g}\left({x}\right)=\left({x}−{p}\right)^{\mathrm{2}} \left({x}−{q}\right)^{\mathrm{2}} \\ $$$${A}=\int_{{p}} ^{{q}} \left[{f}\left({x}\right)−{g}\left({x}\right)\right]{dx} \\ $$$$\:\:\:=\int_{{p}} ^{{q}} \left({x}−{p}\right)^{\mathrm{2}} \left({x}−{q}\right)^{\mathrm{2}} {dx} \\ $$$$\:\:\:=\int_{\mathrm{0}} ^{{q}−{p}} {u}^{\mathrm{2}} \left({u}−\left({q}−{p}\right)\right)^{\mathrm{2}} {du}\:\:{with}\:{u}={x}−{q} \\ $$$$\:\:\:=\int_{\mathrm{0}} ^{{s}} {u}^{\mathrm{2}} \left({u}−{s}\right)^{\mathrm{2}} {du}\:\:{with}\:{s}={q}−{p} \\ $$$$\:\:\:=\int_{\mathrm{0}} ^{{s}} \left({u}^{\mathrm{4}} −\mathrm{2}{su}^{\mathrm{3}} +{s}^{\mathrm{2}} {u}^{\mathrm{2}} \right){du} \\ $$$$\:\:\:=\frac{{s}^{\mathrm{5}} }{\mathrm{5}}−\frac{{s}^{\mathrm{5}} }{\mathrm{2}}+\frac{{s}^{\mathrm{5}} }{\mathrm{3}} \\ $$$$\:\:\:=\frac{{s}^{\mathrm{5}} }{\mathrm{30}} \\ $$$${with}\:{p}=\mathrm{3},\:{q}=\mathrm{9} \\ $$$$\Rightarrow{s}=\mathrm{9}−\mathrm{3}=\mathrm{6} \\ $$$$\Rightarrow{A}=\frac{\mathrm{6}^{\mathrm{5}} }{\mathrm{30}}=\frac{\mathrm{1296}}{\mathrm{5}}\:\checkmark \\ $$
Commented by Frix last updated on 01/Feb/24
$$\mathrm{Nice}! \\ $$