Fundamental Theorem of Calculus/Second Part
Theorem
Let $f$ be a real function which is continuous on the closed interval $\closedint a b$.
Then:
- $(1): \quad f$ has a primitive on $\closedint a b$
- $(2): \quad$ If $F$ is any primitive of $f$ on $\closedint a b$, then:
- $\ds \int_a^b \map f t \rd t = \map F b - \map F a = \bigintlimits {\map F t} a b$
Proof 1
Let $G$ be defined on $\closedint a b$ by:
- $\ds \map G x = \int_a^x \map f t \rd t$
We have:
- $\ds \map G a = \int_a^a \map f t \rd t = 0$ from Integral on Zero Interval
- $\ds \map G b = \int_a^b \map f t \rd t$ from the definition of $G$ above.
Therefore, we can compute:
\(\ds \int_a^b \map f t \rd t\) | \(=\) | \(\ds \int_a^a \map f t \rd t + \int_a^b \map f t \rd t\) | Sum of Integrals on Adjacent Intervals for Continuous Functions | |||||||||||
\(\ds \) | \(=\) | \(\ds \map G a + \map G b\) | From the definition of $G$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \map G b - \map G a\) | $\map G a = 0$ |
By the first part of the theorem, $G$ is a primitive of $f$ on $\closedint a b$.
By Primitives which Differ by Constant‎, we have that any primitive $F$ of $f$ on $\closedint a b$ satisfies $\map F x = \map G x + c$, where $c$ is a constant.
Thus we conclude:
\(\ds \int_a^b \map f t \rd t\) | \(=\) | \(\ds \paren {\map G b + c} - \paren {\map G a + c}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map F b - \map F a\) |
$\blacksquare$
Proof 2
As $f$ is continuous, by the first part of the theorem, it has a primitive. Call it $F$.
$\closedint a b$ can be divided into any number of closed subintervals of the form $\closedint {x_{k - 1} } {x_k}$ where:
- $a = x_0 < x_1 \cdots < x_{k-1} < x_k = b$
Fix such a finite subdivision of the interval $\closedint a b$; call it $P$.
Next, we observe the following telescoping sum identity:
\(\text {(1)}: \quad\) | \(\ds \sum_{i \mathop = 1}^k \paren {\map F {x_i} - \map F {x_{i - 1} } }\) | \(=\) | \(\ds \map F b - \map F a\) |
Because $F' = f$, $F$ is differentiable.
By Differentiable Function is Continuous, $F$ is also continuous.
Therefore we can apply the Mean Value Theorem on $F$.
It follows that in every closed subinterval $I_i = \closedint {x_{i - 1} } {x_i}$ there is some $c_i$ such that:
- $\map {F'} {c_i} = \dfrac {\map F {x_i} - \map F {x_{i - 1} } } {x_i - x_{i - 1} }$
It follows that:
\(\ds \map F {x_i} - \map F {x_{i - 1} }\) | \(=\) | \(\ds \map {F'} {c_i} \paren {x_i - x_{i - 1} }\) | ||||||||||||
\(\text {(2)}: \quad\) | \(\ds \map F b - \map F z\) | \(=\) | \(\ds \sum_{i \mathop = 1}^k \map {F'} {c_i} \paren {x_i - x_{i - 1} }\) | from $(1)$ | ||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{i \mathop = 1}^k \map f {c_i} \paren {x_i - x_{i - 1} }\) | because $F' = f$ |
From the definitions of supremum and infimum, we have for all $i$ (recall $I_i = \closedint {x_{i - 1} } {x_i}$):
- $\displaystyle \inf_{x \mathop \in I_i} \map f x \le \map f {c_i} \le \sup_{x \mathop \in I_i} \map f x$
From the definitions of upper and lower sums, we conclude for any finite subdivision $P$:
- $\displaystyle \map L P \le \sum_{i \mathop = 1}^k \map f {c_i} \paren {x_i - x_{i - 1} } \le \map U P$
Lastly, from the definition of a definite integral and from $(2)$, we conclude:
- $\displaystyle \map F b - \map F a = \int_a^b \map f t \rd t$
$\blacksquare$
Also presented as
The Fundamental Theorem of Calculus can also be presented in the following form:
Let $f$ be a real function which is continuous on the closed interval $\closedint a b$ expressed as a normal first order ODE:
- $(1): \quad \dfrac {\d y} {\d x} = \map f x$
Let $c \in \R$ be an arbitrary real number.
Then there exists a unique solution $\map F x$ to $(1)$ on $\closedint a b$ such that $\map F a = c$, given by the definite integral:
- $\ds \map F x = c + \int_a^x \map f t \rd t$
Sources
- 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text {II}$. Calculus: Definite Integrals
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 15$: Definition of a Definite Integral: $15.2$