Real Analysis/Fundamental Theorem of Calculus

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Real Analysis
Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is said to be the central theorem of elementary calculus. It states effectively that "Differentiation" and "Integration" are inverse operations.

Conventionally, the theorem is presented in two parts

First Form of the Fundamental Theorem[edit]


Let f,F:[a,b]\to\mathbb{R}

Let F be differentiable on [a,b] and let F'(x)=f(x) for all x\in [a,b]

Let f be Riemann integrable on [a,b]

Then, \int_a^b f(x)dx=F(b)-F(a)


Let \int_a^b f=L and let \varepsilon>0 be given.

Then, there exists \delta>0 such that for a partition \|\mathcal{\dot{P}}\|<\delta implies that |S(f,\mathcal{\dot{P}})-L|<\varepsilon

Consider a partition \mathcal{P} and let x_{i-1},x_i\in\mathcal{P}. By Lagrange's Mean Value Theorem, we have that there exists c_i\in (x_{i-1},x_i) that satisfies \frac{F(x_i)-F(x_{i-1})}{x_i-x_{i-1}}=F'(c_i)=f(c_i)

Let the tagged partition \mathcal{\dot{P}} be the partition \mathcal{P} along with the tags c_i

Thus, S(f,\mathcal{\dot{P}})=\sum_{i=1}^n f(c_i)(x_i-x_{i-1})=\sum_{i=1}^n \tfrac{F(x_i)-F(x_{i-1})}{x_i-x_{i-1}}(x_i-x_{i-1})=\sum_{i=1}^n (F(x_i)-F(x_{i-1}))=F(b)-F(a)

But we know that |S(f,\mathcal{\dot{P}})-L|<\varepsilon and hence, |F(b)-F(a)-L|<\varepsilon.

As \varepsilon>0 is arbitrary, |F(b)-F(a)-L|=0 that is, \int_a^b f(x)dx=F(b)-F(a)

Second Form of the Fundamental Theorem[edit]

We first define what is known as the "Indefinite Integral"


Let f:[a,b]\to\mathbb{R} be Riemann integrable on [a,b].

We define the Indefinite Integral of f to be the function F:[a,b]\to\mathbb{R} given by

F(x)=\int_a^x f for all x\in [a,b]


Let f:[a,b]\to\mathbb{R} be continuous at c\in [a,b]

Let F:[a,b]\to\mathbb{R} be the indefinite integral of f

Then, F is differentiable at c and F'(c)=f(c)


Let \varepsilon>0 be given, and let x\in [a,b] but x\neq c

Observe that F(x)-F(c)=\int_c^x f=L(say).

There exists \delta>0 such that if a partition \|\mathcal{\dot{P}}\|<\delta then, |S(f,\mathcal{\dot{P}})-L|<\varepsilon |x-c| (note that in this proof, all the Riemann sums are over the interval [c,x]).

As f is integrable over [c,x], it is bounded over that interval. Hence, let M=\sup f([c,x]). Thus, S(f,\mathcal{\dot{P}})\leq \sum_{i=1}^n M(x_i-x_{i-1})=M(x-c)

As f is continuous at c, there exists \delta>0 such that M(x-c)-S(f,\mathcal{\dot{P}})<\varepsilon |x-c| whenever |x-c|<\delta.

Now consider x\in V_{\delta}(c)

Then, \left| \tfrac{F(x)-F(c)}{x-c}-f(c)\right| =\left| \tfrac{L}{x-c}-f(c)\right| <\left| \tfrac{S(f,\mathcal{\dot{P}})+\varepsilon|x-c|}{x-c}-f(c)\right|<\left| \tfrac{M(x-c)+2\varepsilon|x-c|}{x-c}-f(c)\right|

<\left| \tfrac{f(c)(x-c)+3\varepsilon|x-c|}{x-c}-f(c)\right|<3\varepsilon

That is, \lim_{x\to c}\frac{F(x)-F(c)}{x-c}=f(c), or F'(c)=f(c)