d/dx(e^x) = e^xd/dx(e^{g(x)}) = g'(x)e^{g(x)}(1/2)M(0) = M(0)e^{-kT}M(0): 1/2 = e^{-kT}ln(1/2) = -kTk = ln(2)/T. With T=0.704, kโ0.9846.6.5 ร 0.704 = 4.58 billion yearsL(1) โ 644 gL(2) โ 884 gM(1) โ 1024 โ 644 โ 380 gโซโยน v(t) dt is the total amount of uranium that disappeared (and became lead) from 0 to 1 billion years โ i.e., the increase in L(t) over that interval L(1)=644.
M(1)= Uranium remaining from chart 380 grams - Notice M(1) + L(1) = M(0) = 1024 grams, so the area under v(t) from 0 to 1 must be L(1) = 644 grams.