Fundamental Theorem of Calculus Part Two

Theorem: abf(x)dx=f(b)f(a)(Assuming f(x) is continuous)\text{Theorem:}\space \boxed{\int _{ a }^{ b }{ f\prime \left( x \right) dx } =f\left( b \right) -f\left( a \right)}\quad \text{(Assuming}\space f\prime \left( x \right) \space \text{is continuous)}

 \space

Riemann Integration: fix n and chop [a,b] into n pieces, each piece has length ban=δx\text{Riemann Integration: fix}\space n\space \text{and chop}\space \left[ a,b \right] \space \text{into}\space n\space \text{pieces, each piece has length}\space \frac{b-a}{n}=\delta x

Notation: x0=a, x1=a+δx, x2=a+2δx, , xi=a+iδx, xn=a+nδx=a+n(ban)=a+ba=b\text{Notation:}\space x_0=a,\space x_1=a+\delta x,\space x_2=a+2\delta x,\space \cdots,\space x_i=a+i\delta x,\space x_n=a+n\delta x=a+n\left(\frac{b-a}{n}\right)=a+b-a=b

On each piece [xi1,xi] , choose a point xi , and consider the rectangle of height f(xi) and of width δx\text{On each piece}\space \left[ x_{i-1},x_i \right]\space \text{, choose a point}\space { x }_{ i }^{ \ast }\space \text{, and consider the rectangle of height}\space f\prime\left({ x }_{ i }^{ \ast }\right)\space \text{and of width}\space \delta x

Area of rectangle=(δx)f(xi)Take the sum of the areas of all the rectangles\text{Area of rectangle}=\left(\delta x\right)f\prime\left({ x }_{ i }^{ \ast }\right)\quad \text{Take the sum of the areas of all the rectangles}

limni=1n(δx)f(xi)=defabf(x)dx\lim _{ n\rightarrow \infty }{ \sum _{ i=1 }^{ n }{ \left( \delta x \right) f\prime \left( { x }_{ i }^{ \ast } \right) } } \overset { \text{def} }{ = } \int _{ a }^{ b }{ f\prime \left( x \right) dx }

abf(x)dx=limni=1n(δx)f(xi)\int _{ a }^{ b }{ f\prime \left( x \right) dx }=\lim _{ n\rightarrow \infty }{ \sum _{ i=1 }^{ n }{ \left( \delta x \right) f\prime \left( { x }_{ i }^{ \ast } \right) } }

 \space

By Mean Value Theorem, there is c in (xi1,xi) such that f(xi)f(xi1)xixi1=f(c)\text{By Mean Value Theorem, there is}\space c\space \text{in}\space \left(x_{i-1},x_i \right)\space \text{such that}\space \frac { f\left( { x }_{ i } \right) -f\left( { x }_{ i-1 } \right) }{ { x }_{ i }-{ x }_{ i-1 } } =f\prime \left( c \right)

f(xi)f(xi1)δx=f(c)Let xi=c , then:\frac { f\left( { x }_{ i } \right) -f\left( { x }_{ i-1 } \right) }{ \delta x } =f\prime \left( c \right)\quad \text{Let}\space { x }_{ i }^{ \ast }=c\space \text{, then:}

f(xi)f(xi1)δx=f(xi)\frac { f\left( { x }_{ i } \right) -f\left( { x }_{ i-1 } \right) }{ \delta x } =f\prime \left( { x }_{ i }^{ \ast } \right)

(δx)f(xi)=f(xi)f(xi1)\left( \delta x \right) f\prime \left( { x }_{ i }^{ \ast } \right) =f\left( { x }_{ i } \right) -f\left( { x }_{ i-1 } \right)

 \space

abf(x)dx=limni=1n(δx)f(xi)=limni=1n(f(xi)f(xi1))=limn(f(x1)f(x0)+f(x2)f(x1)++f(xn1)f(xn2)+f(xn)f(xn1))=limn(f(xn)f(x0))=limn(f(b)f(a))abf(x)dx=f(b)f(a)\begin{aligned} \int _{ a }^{ b }{ f\prime \left( x \right) dx } & = \lim _{ n\rightarrow \infty }{ \sum _{ i=1 }^{ n }{ \left( \delta x \right) f\prime \left( { x }_{ i }^{ \ast } \right) } } \\ \quad & = \lim _{ n\rightarrow \infty }{ \sum _{ i=1 }^{ n }{ \left( f\left( { x }_{ i } \right) -f\left( { x }_{ i-1 } \right) \right) } } \\ \quad & = \lim _{ n\rightarrow \infty }{ \left( f\left( { x }_{ 1 } \right) -f\left( { x }_{ 0 } \right) +f\left( { x }_{ 2 } \right) -f\left( { x }_{ 1 } \right) +\cdots +f\left( { x }_{ n-1 } \right) -f\left( { x }_{ n-2 } \right) +f\left( { x }_{ n } \right) -f\left( { x }_{ n-1 } \right) \right) } \\ \quad & = \lim _{ n\rightarrow \infty }{ \left( f\left( { x }_{ n } \right) -f\left( { x }_{ 0 } \right) \right) } \\ \quad & = \lim _{ n\rightarrow \infty }{ \left( f\left( b \right) -f\left( a \right) \right) } \\ \int _{ a }^{ b }{ f\prime \left( x \right) dx } & = f\left( b \right) -f\left( a \right)\quad ∎ \end{aligned}

Note by Gordon Chan
1 month, 1 week ago

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