Fundamental Theorem of Calculus Part Two

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

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

$\text{Notation:}\space x_0=a,\space x_1=a+\delta x,\space x_2=a+2\delta x,\space \cdots,\space x_i=a+i\delta x,\space x_n=a+n\delta x=a+n\left(\frac{b-a}{n}\right)=a+b-a=b$

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

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

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

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

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

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

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

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

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$\begin{aligned} \int _{ a }^{ b }{ f\prime \left( x \right) dx } & = \lim _{ n\rightarrow \infty }{ \sum _{ i=1 }^{ n }{ \left( \delta x \right) f\prime \left( { x }_{ i }^{ \ast } \right) } } \\ \quad & = \lim _{ n\rightarrow \infty }{ \sum _{ i=1 }^{ n }{ \left( f\left( { x }_{ i } \right) -f\left( { x }_{ i-1 } \right) \right) } } \\ \quad & = \lim _{ n\rightarrow \infty }{ \left( f\left( { x }_{ 1 } \right) -f\left( { x }_{ 0 } \right) +f\left( { x }_{ 2 } \right) -f\left( { x }_{ 1 } \right) +\cdots +f\left( { x }_{ n-1 } \right) -f\left( { x }_{ n-2 } \right) +f\left( { x }_{ n } \right) -f\left( { x }_{ n-1 } \right) \right) } \\ \quad & = \lim _{ n\rightarrow \infty }{ \left( f\left( { x }_{ n } \right) -f\left( { x }_{ 0 } \right) \right) } \\ \quad & = \lim _{ n\rightarrow \infty }{ \left( f\left( b \right) -f\left( a \right) \right) } \\ \int _{ a }^{ b }{ f\prime \left( x \right) dx } & = f\left( b \right) -f\left( a \right)\quad ∎ \end{aligned}$