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Question Number 33867 by Joel578 last updated on 26/Apr/18

Given A(t) is an area bounded between y = x^2 + tx and x−axis, 0 < t < 2 Find the propability we choose t so (1/(48)) ≤ A(t) ≤ (1/(16))

GivenA(t)isanareaboundedbetweeny=x2+txandx−axis,0<t<2Findthepropabilitywechoosetso148⩽A(t)⩽116

Answered by MJS last updated on 26/Apr/18

x^2 +tx=0 x(x+t)=0 x_1 =0 x_2 =−t ∣∫_(−t) ^0 (x^2 +tx)dx∣=∣[(x^3 /3)+((tx^2 )/2)]_(−t) ^0 ∣=∣(t^3 /3)−(t^3 /2)∣=(t^3 /6) (1/(48))≤(t^3 /6)≤(1/(16)) (1/8)≤t^3 ≤(3/8) (1/2)≤t≤((3)^(1/3) /2) ((((3)^(1/3) /2)−(1/2))/2)=(((3)^(1/3) −1)/4)≈.1106 ⇒ 11.06%

x2+tx=0x(x+t)=0x1=0x2=−tMissing \left or extra \right148⩽t36⩽11618⩽t3⩽3812⩽t⩽332332−122=33−14≈.1106⇒11.06%

Commented by Joel578 last updated on 26/Apr/18

thank you very much

thankyouverymuch

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