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Question Number 88616 by TawaTawa1 last updated on 11/Apr/20

Commented by jagoll last updated on 28/Apr/20

1.x^2 −4x+0 = 0 a = 1 , b = −4 , c = 0

$$\mathrm{1}.{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{0}\:=\:\mathrm{0} \\ $$$${a}\:=\:\mathrm{1}\:,\:{b}\:=\:−\mathrm{4}\:,\:{c}\:=\:\mathrm{0} \\ $$

Commented by ajfour last updated on 11/Apr/20

Commented by ajfour last updated on 11/Apr/20

eq. of line AB you should write first. y−9=(((1−9)/(4−0)))(x−0) ⇒ y=9−2x trapezium area ABFO is ∫_0 ^( 4) (9−2x)dx (blue+black area) black area=∫_0 ^( 4) (x−3)^2 dx blue area=(blue+black)−(black) = ∫_0 ^( 4) (9−2x)dx−∫_0 ^( 4) (x−3)^2 = {9x−x^2 −(((x−3)^3 )/3)}∣_0 ^( 4) =( 36−16−(1/3))−(−(((−3)^3 )/3)) = 20−(1/3)−9 = ((32)/3) . this is the simple way.

$${eq}.\:{of}\:{line}\:{AB}\:{you}\:{should}\:{write} \\ $$$${first}. \\ $$$${y}−\mathrm{9}=\left(\frac{\mathrm{1}−\mathrm{9}}{\mathrm{4}−\mathrm{0}}\right)\left({x}−\mathrm{0}\right) \\ $$$$\Rightarrow\:\:{y}=\mathrm{9}−\mathrm{2}{x} \\ $$$${trapezium}\:{area}\:{ABFO}\:\:{is} \\ $$$$\:\:\:\int_{\mathrm{0}} ^{\:\mathrm{4}} \left(\mathrm{9}−\mathrm{2}{x}\right){dx}\:\:\:\:\left({blue}+{black}\:{area}\right) \\ $$$${black}\:{area}=\int_{\mathrm{0}} ^{\:\:\mathrm{4}} \left({x}−\mathrm{3}\right)^{\mathrm{2}} {dx} \\ $$$${blue}\:{area}=\left({blue}+{black}\right)−\left({black}\right) \\ $$$$\:\:\:\:\:=\:\int_{\mathrm{0}} ^{\:\mathrm{4}} \left(\mathrm{9}−\mathrm{2}{x}\right){dx}−\int_{\mathrm{0}} ^{\:\mathrm{4}} \left({x}−\mathrm{3}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:=\:\left\{\mathrm{9}{x}−{x}^{\mathrm{2}} −\frac{\left({x}−\mathrm{3}\right)^{\mathrm{3}} }{\mathrm{3}}\right\}\mid_{\mathrm{0}} ^{\:\mathrm{4}} \\ $$$$\:\:\:=\left(\:\mathrm{36}−\mathrm{16}−\frac{\mathrm{1}}{\mathrm{3}}\right)−\left(−\frac{\left(−\mathrm{3}\right)^{\mathrm{3}} }{\mathrm{3}}\right) \\ $$$$\:\:\:=\:\mathrm{20}−\frac{\mathrm{1}}{\mathrm{3}}−\mathrm{9}\:=\:\frac{\mathrm{32}}{\mathrm{3}}\:. \\ $$$${this}\:{is}\:{the}\:{simple}\:{way}. \\ $$

Commented by TawaTawa1 last updated on 11/Apr/20

Wow, God bless you sir, but sir mrW got ((32 )/3)

$$\mathrm{Wow},\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir},\:\:\mathrm{but}\:\mathrm{sir}\:\:\mathrm{mrW}\:\:\mathrm{got}\:\:\:\frac{\mathrm{32}\:}{\mathrm{3}}\:\: \\ $$

Commented by I want to learn more last updated on 11/Apr/20

I understand it sir very well

$$\mathrm{I}\:\mathrm{understand}\:\mathrm{it}\:\mathrm{sir}\:\mathrm{very}\:\mathrm{well} \\ $$

Commented by TawaTawa1 last updated on 11/Apr/20

God bless you sir. It is clear.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{It}\:\mathrm{is}\:\mathrm{clear}. \\ $$

Commented by jagoll last updated on 12/Apr/20

the equation of line determinant (((0 9)),((4 1)))⇒ (9−1)x+(4−0)y=36 8x +4y = 36 ⇒y=9−2x the parabolic line cut ⇒ (x−3)^2 =9−2x x^2 −4x=0 , Δ = 16 area = ((Δ(√Δ))/(6.a^2 )) = ((16.4)/(6.1))=((32)/3)

$$\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{line}\: \\ $$$$\begin{vmatrix}{\mathrm{0}\:\:\:\mathrm{9}}\\{\mathrm{4}\:\:\:\mathrm{1}}\end{vmatrix}\Rightarrow\:\left(\mathrm{9}−\mathrm{1}\right)\mathrm{x}+\left(\mathrm{4}−\mathrm{0}\right)\mathrm{y}=\mathrm{36} \\ $$$$\mathrm{8x}\:+\mathrm{4y}\:=\:\mathrm{36}\:\Rightarrow\mathrm{y}=\mathrm{9}−\mathrm{2x} \\ $$$$\mathrm{the}\:\mathrm{parabolic}\:\mathrm{line}\:\mathrm{cut}\:\Rightarrow\:\left(\mathrm{x}−\mathrm{3}\right)^{\mathrm{2}} =\mathrm{9}−\mathrm{2x} \\ $$$$\mathrm{x}^{\mathrm{2}} −\mathrm{4x}=\mathrm{0}\:,\:\Delta\:=\:\mathrm{16}\: \\ $$$$\mathrm{area}\:=\:\frac{\Delta\sqrt{\Delta}}{\mathrm{6}.\mathrm{a}^{\mathrm{2}} }\:=\:\frac{\mathrm{16}.\mathrm{4}}{\mathrm{6}.\mathrm{1}}=\frac{\mathrm{32}}{\mathrm{3}} \\ $$

Commented by TawaTawa1 last updated on 12/Apr/20

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

Commented by I want to learn more last updated on 17/Apr/20

How is a = 1 sir

$$\mathrm{How}\:\mathrm{is}\:\:\mathrm{a}\:\:=\:\:\mathrm{1}\:\:\mathrm{sir} \\ $$

Answered by mr W last updated on 11/Apr/20

(2/3)×[((9+1)/2)−(2−3)^2 ]×4=((32)/3)

$$\frac{\mathrm{2}}{\mathrm{3}}×\left[\frac{\mathrm{9}+\mathrm{1}}{\mathrm{2}}−\left(\mathrm{2}−\mathrm{3}\right)^{\mathrm{2}} \right]×\mathrm{4}=\frac{\mathrm{32}}{\mathrm{3}} \\ $$

Commented by I want to learn more last updated on 11/Apr/20

Sir, i understand ∫_0 ^( 4) (x − 3)^2 dx = ((28)/3) But how can i understand the other part.

$$\mathrm{Sir},\:\mathrm{i}\:\mathrm{understand}\:\:\:\:\int_{\mathrm{0}} ^{\:\mathrm{4}} \:\:\left(\mathrm{x}\:−\:\mathrm{3}\right)^{\mathrm{2}} \:\:\mathrm{dx}\:\:\:\:=\:\:\:\frac{\mathrm{28}}{\mathrm{3}} \\ $$$$\mathrm{But}\:\mathrm{how}\:\mathrm{can}\:\mathrm{i}\:\mathrm{understand}\:\mathrm{the}\:\mathrm{other}\:\mathrm{part}. \\ $$

Commented by TawaTawa1 last updated on 11/Apr/20

God bless you sir, i also have the same question. Same question ′i want to learn asked′

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir},\:\:\mathrm{i}\:\mathrm{also}\:\mathrm{have}\:\mathrm{the}\:\mathrm{same}\:\mathrm{question}. \\ $$$$\mathrm{Same}\:\mathrm{question}\:\:'\mathrm{i}\:\mathrm{want}\:\mathrm{to}\:\mathrm{learn}\:\mathrm{asked}' \\ $$

Commented by mr W last updated on 12/Apr/20

see Q88758

$${see}\:{Q}\mathrm{88758} \\ $$

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