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Question-209659




Question Number 209659 by mokys last updated on 17/Jul/24
Answered by efronzo1 last updated on 18/Jul/24
   S = ∫_0 ^1  2^x  dx + ∫_1 ^(√3)  (3−x^2 )dx
S=102xdx+31(3x2)dx
Answered by mr W last updated on 18/Jul/24
Commented by mr W last updated on 18/Jul/24
T=∫_0 ^1 (3−x^2 −2^x )dx     =[3x−(x^3 /3)−(2^x /(ln 2))]_0 ^1      =3−(1/3)−((2−1)/(ln 2))=(8/3)−(1/(ln 2))  S+T=(2/3)×3×(√3)=2(√3)  ⇒S=2(√3)−(8/3)+(1/(ln 2))≈2.24
T=01(3x22x)dx=[3xx332xln2]01=31321ln2=831ln2S+T=23×3×3=23S=2383+1ln22.24
Answered by lepuissantcedricjunior last updated on 18/Jul/24
f(x)=3−x^2 ;g(x)=2^x   2^x +x^2 −3=0=>x=1=>0≤x≤1  S=∫_0 ^1 ∫_0 ^2^x  dx=[(1/(ln2))2^x ]_0 ^1 =(1/(ln2))≈(1/(0.69))≈((100)/(69))
f(x)=3x2;g(x)=2x2x+x23=0=>x=1=>0x1S=0102xdx=[1ln22x]01=1ln210.6910069

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