Infi-2 6

Mean value theorem for definite integrals #theorem

Let f be continuous on [a,b]Then ∃c∈[a,b]:∫abf(x)dx=f(c)(b−a)

Definite Integral in the point #definition

∫aaf(x)dx=0

Definite integral on inverse interval #theorem

∫abf(x)dx=−∫baf(x)dx

Area function #definition

Let f be Riemann-integrable on [a,b]S is called an area function of f and defined asS(x)=∫axf(t)dtLet f(x)=x on [0,1]⟹S(x)=x22S′(x)=f(x)

Continuity of area function #theorem

Let f be a Riemann-integrable function on [a,b]Let S be an area function of fThen S is continuousProof:Let c∈[a,b]Let x→c0≤|S(x)−S(c)|=|∫axf(t)dt−∫acf(t)dt|∫axf(t)dt=∫acf(t)d(t)+∫cxf(t)dt⟹|∫axf(t)dt−∫acf(t)dt|=|∫cxf(t)dt|≤∫cx|f(t)|dtf is Riemann-integrable⟹|f| is Riemann-integrable⟹|f| is bounded⟹∃M:|f|≤M⟹∫cx|f(t)|dt≤∫cxMdt=M(x−c)x→c⟹x−c→00≤|S(x)−S(c)|≤M(x−c)→0⟹S(x)−S(c)→0⟹limx→cS(x)=S(c)⟹S is continuous on [a,b]

Fundamental theorem of Calculus (Part 1) #theorem

Let f be continuous on [a,b]Then S(x)=∫axf(t)dt is differentiable and S′(x)=f(x)Proof:Let c∈[a,b]S′(c)=limh→0S(c+h)−S(c)h=limh→0(∫ac+hf(t)dt−∫acf(t)dt)h==limh→0∫cc+hf(t)dthf is continuous ⟹∃d∈[c,c+h]:∫cc+hf(t)dt=f(d)(c+h−c)=f(d)h⟹limh→0∫cc+hf(t)dth=limh→0f(d)hh=limh→0f(d)c≤d≤c+h⏟→c+0=c⟹d→cf is continuous ⟹f(d)→f(c)⟹S′(c)=limh→0f(d)=f(c)

Fundamental theorem of Calculus (Part 2) aka Newton-Leibniz theorem #theorem

Let f be Riemann-integrable on [a,b]Let F be continuous and a primitive of fThen ∫abf(x)dx=F(b)−F(a)Proof:Let f be continuousF′−S′=f−f=0⟹∃C:∀x∈[a,b]:F(x)=S(x)+C∫abf(t)dt=∫abf(t)dt−0=∫abf(t)dt−∫aaf(t)dt==S(b)−S(a)=F(b)+C−F(a)−C=F(b)−F(a)Let f be non-continuousF is continuous on [a,b] and differentiable on (a,b)⟹∃c∈(a,b):F′(c)(b−a)=F(b)−F(a)Let a=x0<x1<⋯<xn=bF(b)−F(a)=F(xn)−F(x0)==F(xn)+(−F(xn−1)+F(xn−1))+⋯+(−F(x1)+F(x1))−F(x0)==(F(xn)−F(xn−1))+(F(xn−1)−F(xn−2))+⋯+(F(x2)−F(x1))+(F(x1)−F(x0))F(b)−F(a)=∑i=1n(F(xi)−F(xi−1))∀i∈[1,n]:∃ci∈[xi−1,xi]:F(xi)−F(xi−1)=F′(ci)(xi−xi−1)⟹∑i=1n(F(xi)−F(xi−1))=∑i=1nf(ci)Δxi=S(f,P,C)Where P={x0,x1,…,xn}C={c1,c2,…,cn}⟹limλ(P)→0F(b)−F(a)=limλ(P)→0S(f,P,C)⟹F(b)−F(a)=∫abf(x)dx
The following notation can be used: F(b)−F(a)=F(x)|x=ax=b
Let f(x)={10≤x≤101<x≤2 on [0,2]f is Riemann-integrable by Lebesgue criterion, it is bounded and has one discontinuityBut, by Darboux theorem, there is no primitive of f because it has a jump discontinuityEven more that that, if f has a primitive, it does not imply that f is Riemann-integrableF(x)={x2sin⁡(1x3)x≠00x=0F′(0)=limh→0F(h)−F(0)h=limh→0F(h)h=limh→0hsin⁡(1h3)=0x≠0⟹F′(x)=2xsin⁡(1x3)+(−3x2)cos⁡(1x3)⟹F′(x)=f(x)={2xsin⁡(1x3)−3x2cos⁡(1x3)x≠00x=0f is not bounded because of 3x2, when x→0⟹f is not Riemann-integrable, but it does have a primitive−F(x)
limx→0∫0xsin⁡(t2)dtx3=???