Fundamental Theorem of Calculus/Second Part
Contents |
Theorem
Let $f$ be a real function which is continuous on the closed interval $\left[{a .. b}\right]$.
Then:
- $f$ has a primitive on $\left[{a .. b}\right]$
- If $F$ is any primitive of $f$ on $\left[{a .. b}\right]$, then:
- $\displaystyle \int_a^b f \left({t}\right) \ \mathrm d t = F \left({b}\right) - F \left({a}\right) = \left[{ F \left({t}\right) }\right]_a^b$
Proof 1
Let $G$ be defined on $\left[{a .. b}\right]$ by:
- $\displaystyle G \left({x}\right) = \int_a^x f \left({t}\right) \ \mathrm d t$
We have:
- $\displaystyle G \left({a}\right) = \int_a^a f \left({t}\right) \ \mathrm d t = 0$ from Integral on Zero Interval
- $\displaystyle G \left({b}\right) = \int_a^b f \left({t}\right) \ \mathrm d t$ from the definition of $G$ above.
Therefore, we can compute:
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \int_a^b f \left({t}\right) \ \mathrm d t\) | \(=\) | \(\displaystyle \int_a^a f \left({t}\right) \ \mathrm d t + \int_a^b f \left({t}\right) \ \mathrm d t\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Sum of Integrals on Adjacent Intervals | ||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle G \left({a}\right) + G \left({b}\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | From the definition of $G$ | ||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle G \left({b}\right) - G \left({a}\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | $G \left({a}\right) = 0$ |
By the first part of the theorem, $G$ is a primitive of $f$ on $\left[{a .. b}\right]$.
By Primitives which Differ by a Constant‎, we have that any primitive $F$ of $f$ on $\left[{a .. b}\right]$ satisfies $F \left({x}\right) = G \left({x}\right) + c$, where $c$ is a constant.
Thus we conclude:
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \int_a^b f \left({t}\right) \ \mathrm d t\) | \(=\) | \(\displaystyle \left({G \left({b}\right) + c}\right) - \left({G \left({a}\right) + c}\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle F \left({b}\right) - F \left({a}\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) |
$\blacksquare$
Proof 2
As $f$ is continuous, by the first part of the theorem, it has a primitive. Call it $F$.
$\left[{a .. b}\right]$ can be divided into any number of closed subintervals of the form $\left[{x_{k-1} .. x_k}\right]$ where:
- $ a = x_0 < x_1 \cdots < x_{k-1} < x_k = b$
Fix such a subdivision of the interval $\left[{a .. b}\right]$; call it $P$.
Next, we observe the following telescoping sum identity:
| \((1):\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \sum_{i=1}^{k}\ F \left({x_i}\right) - F \left({ x_{i-1} }\right)\) | \(=\) | \(\displaystyle F \left({b}\right) - F \left({a}\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) |
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 $I_i = \left[{x_{i-1} .. x_i}\right]$ there is some $c_i$ such that:
- $F' \left({c_i}\right) = \dfrac {F \left({x_i}\right) - F \left({x_{i-1}}\right)} {x_{i} - x_{i-1}}$
It follows that:
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle F \left({x_i}\right) - F \left({x_{i-1} }\right)\) | \(=\) | \(\displaystyle F' \left({c_i}\right) \left({ x_{i} - x_{i-1} }\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \((2):\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle F \left({b}\right) - F \left({a}\right)\) | \(=\) | \(\displaystyle \sum_{i=1}^k F' \left({c_i}\right) \left({ x_{i} - x_{i-1} }\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | By equation $\left({1}\right)$ | |
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \sum_{i=1}^{k} f \left({c_i}\right) \left({ x_{i} - x_{i-1} }\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Because $F' = f$ |
From the definitions of supremum and infimum, we have for all $i$ (recall $I_i = \left[{x_{i-1} .. x_i}\right]$):
- $\displaystyle \inf_{x \in I_i} \ f \left({x}\right) \le f \left({c_i}\right) \le \sup_{x \in I_i} \ f \left({x}\right)$
From the definitions of upper and lower sums, we conclude for any subdivision $P$:
- $\displaystyle L \left({P}\right) \le \sum_{i=1}^{k} f \left({c_i}\right) \left({ x_{i} - x_{i-1} }\right) \le U \left({P}\right)$
Lastly, from the definition of a definite integral and from $\left({2}\right)$, we conclude:
- $\displaystyle F \left({b}\right) - F \left({a}\right) = \int_a^b f \left({t}\right) \ \mathrm d t$
$\blacksquare$
Sources
- K.G. Binmore: Mathematical Analysis: A Straightforward Approach (1977)... (previous)... (next): $\S 13.14$
