Question Number 154142 by EDWIN88 last updated on 14/Sep/21
Answered by mr W last updated on 14/Sep/21
![y=a(x−4)(x+2)^2 8=a(−4)(+2)^2 ⇒a=−(1/2) y=−(1/2)(x−4)(x+2)^2 (dy/dx)=−(1/2)[(x+2)^2 +2(x−4)(x+2)]=0 (x+2)(x−2)=0 ⇒x=−2 (rejected) ⇒x=2 ⇒x_P =2, y_P =−(1/2)(2−4)(2+2)^2 =16 eqn. of OP: y=8x eqn. of QR: y=8−(x/8) 8−(x/8)=8x ⇒x_R =((64)/(65)) A=∫_0 ^4 ydx−(1/2)×8×((64)/(65)) A=−(1/2)∫_0 ^4 (x−4)(x+2)^2 dx−((256)/(65)) A=−(1/2)∫_0 ^4 (x^3 −12x−16)dx−((256)/(65)) A=−(1/2)[(4^4 /4)−6×4^2 −16×4]−((256)/(65)) A=48−((256)/(65))=((2864)/(65))≈44.06](https://www.tinkutara.com/question/Q154147.png)
$${y}={a}\left({x}−\mathrm{4}\right)\left({x}+\mathrm{2}\right)^{\mathrm{2}} \\ $$$$\mathrm{8}={a}\left(−\mathrm{4}\right)\left(+\mathrm{2}\right)^{\mathrm{2}} \:\Rightarrow{a}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${y}=−\frac{\mathrm{1}}{\mathrm{2}}\left({x}−\mathrm{4}\right)\left({x}+\mathrm{2}\right)^{\mathrm{2}} \\ $$$$\frac{{dy}}{{dx}}=−\frac{\mathrm{1}}{\mathrm{2}}\left[\left({x}+\mathrm{2}\right)^{\mathrm{2}} +\mathrm{2}\left({x}−\mathrm{4}\right)\left({x}+\mathrm{2}\right)\right]=\mathrm{0} \\ $$$$\left({x}+\mathrm{2}\right)\left({x}−\mathrm{2}\right)=\mathrm{0} \\ $$$$\Rightarrow{x}=−\mathrm{2}\:\left({rejected}\right) \\ $$$$\Rightarrow{x}=\mathrm{2}\:\Rightarrow{x}_{{P}} =\mathrm{2},\:{y}_{{P}} =−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}−\mathrm{4}\right)\left(\mathrm{2}+\mathrm{2}\right)^{\mathrm{2}} =\mathrm{16} \\ $$$${eqn}.\:{of}\:{OP}: \\ $$$${y}=\mathrm{8}{x} \\ $$$${eqn}.\:{of}\:{QR}: \\ $$$${y}=\mathrm{8}−\frac{{x}}{\mathrm{8}} \\ $$$$\mathrm{8}−\frac{{x}}{\mathrm{8}}=\mathrm{8}{x}\:\Rightarrow{x}_{{R}} =\frac{\mathrm{64}}{\mathrm{65}} \\ $$$${A}=\int_{\mathrm{0}} ^{\mathrm{4}} {ydx}−\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{8}×\frac{\mathrm{64}}{\mathrm{65}} \\ $$$${A}=−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{4}} \left({x}−\mathrm{4}\right)\left({x}+\mathrm{2}\right)^{\mathrm{2}} {dx}−\frac{\mathrm{256}}{\mathrm{65}} \\ $$$${A}=−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{4}} \left({x}^{\mathrm{3}} −\mathrm{12}{x}−\mathrm{16}\right){dx}−\frac{\mathrm{256}}{\mathrm{65}} \\ $$$${A}=−\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\mathrm{4}^{\mathrm{4}} }{\mathrm{4}}−\mathrm{6}×\mathrm{4}^{\mathrm{2}} −\mathrm{16}×\mathrm{4}\right]−\frac{\mathrm{256}}{\mathrm{65}} \\ $$$${A}=\mathrm{48}−\frac{\mathrm{256}}{\mathrm{65}}=\frac{\mathrm{2864}}{\mathrm{65}}\approx\mathrm{44}.\mathrm{06} \\ $$
Commented by iloveisrael last updated on 15/Sep/21
$${yes} \\ $$
Answered by iloveisrael last updated on 15/Sep/21
