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Question Number 128851 by mnjuly1970 last updated on 10/Jan/21
               ...nice   calculus  (I)...     calculate  ::         Ψ=∫_0 ^( 1) {(e−1)(√(log( 1+ex−x ))) +e^x^2  }dx=?
nicecalculus(I)calculate::Ψ=01{(e1)log(1+exx)+ex2}dx=?
Commented by mindispower last updated on 10/Jan/21
{.} fraction part ?
{.}fractionpart?
Commented by mnjuly1970 last updated on 11/Jan/21
no mr power
nomrpower
Answered by Olaf last updated on 11/Jan/21
Let f(x) = (e−1)(√(ln(1+(e−1)x))+e^x^2    f is clearly an increasing function and  f(0) = 1, f(1) = e  ⇒ ∃a∈]0,1[ \ f(a) = 2    Ψ = ∫_0 ^1 {f(x)}dx = ∫_0 ^a {f(x)}dx+∫_a ^1 {f(x)}dx  Ψ = ∫_0 ^a (f(x)−1)dx+∫_a ^1 (f(x)−2)dx  Ψ = ∫_0 ^1 f(x)dx−a−2(1−a)  Ψ = ∫_0 ^1 f(x)dx+a−2  Let u = ln(1+(e−1)x)  e^u  = 1+(e−1)x  e^u du = (e−1)dx    I = (e−1)∫_0 ^1 (√(ln(1+(e−1)x)))dx  I = (e−1)∫_0 ^1 (√u)((e^u du)/(e−1))  I = ∫_0 ^1 (√u)e^u du  I = [(√x)e^x −((√π)/2)erfi((√x))]_0 ^1  = e−((√π)/2)erfi(1)    J = ∫_0 ^1 e^x^2  dx = [((√π)/2)erfi(x)]_0 ^1  = ((√π)/2)erfi(1)    Ψ = I+J+a−2 = e+a−2  (a ≈ 0,542)
Letf(x)=(e1)ln(1+(e1)x+ex2fisclearlyanincreasingfunctionandf(0)=1,f(1)=ea]0,1[f(a)=2Ψ=01{f(x)}dx=0a{f(x)}dx+a1{f(x)}dxΨ=0a(f(x)1)dx+a1(f(x)2)dxΨ=01f(x)dxa2(1a)Ψ=01f(x)dx+a2Letu=ln(1+(e1)x)eu=1+(e1)xeudu=(e1)dxI=(e1)01ln(1+(e1)x)dxI=(e1)01ueudue1I=01ueuduI=[xexπ2erfi(x)]01=eπ2erfi(1)J=01ex2dx=[π2erfi(x)]01=π2erfi(1)Ψ=I+J+a2=e+a2(a0,542)
Commented by mnjuly1970 last updated on 11/Jan/21
grateful for your work mr olaf..
gratefulforyourworkmrolaf..
Answered by mnjuly1970 last updated on 11/Jan/21
   note: ∫_(a ) ^( b) f(x)dx+∫_c ^( d) f^( −1) (x)dx=bd−ac    Ψ=∫_0 ^( 1) (e−1)(√(log(1+x(e−1))) dx              + ∫_0 ^( 1) e^x^2  dx    ∴ Ψ=^([u=(1+x(e−1)]) ∫_1 ^( e) (√(log(u))) du+∫_0 ^( 1) e^x^2  dx           =∫_1 ^( e) (√(log(x))) +∫_0 ^( 1) e^x^2  dx=e      note: f(x)=(√(log(x))) ⇔ f^( −1) (x)=e^x^2
note:abf(x)dx+cdf1(x)dx=bdacΨ=01(e1)log(1+x(e1)dx+01ex2dxΨ=[u=(1+x(e1)]1elog(u)du+01ex2dx=1elog(x)+01ex2dx=enote:f(x)=log(x)f1(x)=ex2
Answered by mnjuly1970 last updated on 11/Jan/21

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