# PharmPK Discussion - How to express AUEC in terms of AUC

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• On 7 May 2009 at 18:08:33, "Jack G. Shi" (JShi.at.incyte.com) sent the message
`The following message was posted to: PharmPKGroup,Can someone post an equation to express the area under effect vs. timecurve (AUEC or AUE) in the term of area under plasma concentrationcurve (AUC)?Assume the PD effect follows a simple Emax/EC50 model:E(t) = Emax*C(t)/(EC50 + C(t)), where C(t) is the PK curve.Thanks,Jack`
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• On 9 May 2009 at 15:19:54, Wojciech Jawien (mfjawien.aaa.cyf-kr.edu.pl) sent the message
`Dear Jack,no one can post such an equation, since in general (i.e. without veryspecific assumptions on C(t)) it does not exist.It cannot exist, because AUE is not a unique function of AUC.To demonstrate this, assume a very simple case:Emax=1EC50=1and consider two concentration-time profiles, such that C2(t)=0.5*C1(t/2) for any t. The only additional assumption on C1 is the AUC1 existat all, besides that the profile may be arbitrary.Using basic principles of integral calculus you find that AUC1=AUC2,but AUE2 > AUE1 (for a detailed proof see below).AUE rarely may be expressed analyticaly by PK parameters:For the simplest equation C(t)=C0*exp(-k*t) one obtains AUC=C0/k andAUE=Emax/k * log(1+C0/EC50). Note that even for such a simple case youcannot calculate AUE based only on AUC value - you need to know alsoC0 or k.For a two-exponential equation (for instance C(t)=A*(exp(-k*t)-exp(-ka*t)) ) probably there is no closed form for AUE. I tried tointegrate it with Wolfram Mathematica 7.0 and (after several minutesof hard work!) the integrating routine gave up.Detailed proof:Assume all integrals are in the interval from 0 to infinity.AUC2=Integral C2(t) dt = Integral 0.5*C1(t/2) dt = 0.5*IntegralC1(u)*2 du = Integral C1(u) du = AUC1   { substitution you = t/2 wasused, thus dt = 2 du }AUE2=Integral C2(t) / [1+C2(t)] dt = Integral 1 / [1 + 1/C2(t)] dt =Integral 1 / [1+1/( 0.5*C1(t/2) )] dt = Integral { 1 / [1+1/( 0.5*C1(u) )]  }*2 du Integral 2 / [1 + 2/C1(u)] du > Integral 2 / [2+ 2/C1(u)] du = Integral 1 / [1 + 1/C1(u)] du = AUE1best regardsWojciech JawienJagiellonian UniversityFaculty of PharmacyKrakow, Poland`
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• On 11 May 2009 at 11:26:21, "Jack G. Shi" (JShi.-at-.incyte.com) sent the message
`The following message was posted to: PharmPKThanks for your detailed math exercise and explain why there could notbe an unique expression of AUE in terms of AUC, IC50 etc.  Icompletely agree that AUE is a function of shape of C(t) and not justAUC.  In hindsight, my real interest was to know how AUE is related todose in the case that PK is perfectly linear, i.e., the shape of C(t)does not change.  I doubt there is a solution even for that specialcase.  But the equation you posted with the simplest 1C IV PK curve ishelpful.  Thanks!Jack`
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