Unlock the Secret to Integrability: A Mathematical Proof Revealed

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\[M_{k}-m_{k}<\frac{\epsilon}{b-a},\]

and so

\[U(f,P)-L(f,P)=\sum_{k=1}^{n}(M_{k}-m_{k})(x_{k}-x_{k-1})<\epsilon,\]

which shows that \(f\) is integrable on \([a,b]\)

**Remark 28**.: The argument used above can be modified to show that if \(f\) is a monotonic bounded function, then it is integrable. Because \(f\) is bounded, there exists \(M>0\) such that \(|f(x)|\leq M\) on \([a,b]\). Given \(\epsilon>0\), choose a partition \(P\) of \(n\) intervals of length less than \(\dfrac{\epsilon}{2M}\). Then

\[U(f,P)-L(f,P)=\sum_{k=1}^{n}(f(x_{k})-f(x_{k-1}))(x_{k}-x_{k-1})<\dfrac{ \epsilon}{2M}[f(b)-f(a)]\leq\epsilon.\]

Hence \(f\) is integrable on \([a,b]\).

### Algebraic Properties of the Integral

Suppose \(f\) and \(g\) are integrable functions on \([a,b]\), and suppose that \(c\) is a constant. Then the functions \(cf,f+g,f\pm g\), and \(|f|\) are integrable on \([a,b]\) and

* \(\int_{a}^{b}cf(x)\,dx=c\int_{a}^{b}f(x)\,dx\).
* \(\int_{a}^{b}[f(x)\pm g(x)]\,dx=\int_{a}^{b}f(x)\,dx\pm\int_{a}^{b}g(x)\,dx\) (additivity).
* \(\left|\int_{a}^{b}f(x)\,dx\right|\leq\int_{a}^{b}|f(x)|\,dx\).

Moreover, if \(f(x)\leq g(x)\) for all \(x\in[a,b]\), then \(\int_{a}^{b}f(x)\,dx\leq\int_{a}^{b}g(x)\,dx\). The above algebraic properties follow fairly readily from the definitions. Note also that additivity of the integral suggests an appropriate extension of integrating complex-valued functions. More precisely if \(f\) is a bounded complex-valued function on \([a,b]\), meaning if \(f(x)=\operatorname{Re}\!f(x)+i\operatorname{Im}\!f(x)\), then

\[\int_{a}^{b}f(x)\,dx=\int_{a}^{b}\operatorname{Re}\!f(x)\,dx+i\int_{a}^{b} \operatorname{Im}\!f(x)\,dx.\]

There is another kind of additivity of the integral stated below, which can be proven by using partitions that include the point \(b\).

**Proposition 17**.: _Suppose that \(a

\[\int_{a}^{c}f(x)\,dx=\int_{a}^{b}f(x)\,dx+\int_{b}^{c}f(x)\,dx.\]

The above identity is quite useful especially if we drop the condition \(a

\[\int_{b}^{a}f(x)\,dx=-\int_{a}^{b}f(x)\,dx.\]

The conclusion of the above proposition is valid for any triple of numbers \(a,b\), and \(c\) for which all three integrals are defined.

### Intermediate Value Theorem for Integrals

The Intermediate Value Theorem that we have seen for continuous functions can be used to obtain a corresponding theorem for integrals.

**Theorem 50**.: _If a function \(f\) is continuous on an interval \([a,b]\), then (Figure 1.48)_

\[\frac{1}{b-a}\int_{a}^{b}f(x)\,dx=f(c)\quad\text{for some $c\in[a,b]$}.\]

The area of the rectangle with height \(f(c)\) and base \((b-a)\) equals the area under \(f(x)\) over \([a,b]\).

Proof

: Since \(f\) is continuous on the compact set \([a,b]\), it attains its minimum value \(m\) and a maximum value \(M\) on the interval \([a,b]\). Since \(m\leq f(x)\leq M\) for all \(x\in[a,b]\), it follows that

\[m\leq\frac{1}{b-a}\int_{a}^{b}f(x)\,dx\leq M.\]

An application of Intermediate Value Theorem for continuous functions gives the desired result.

**Remark 29**.: To extend the Riemann integral to unbounded intervals or unbounded functions, we define

\[\int_{a}^{\infty}f(x)\,dx=\lim_{b\to\infty}\int_{a}^{b}f(x)\,dx\quad\text{and }\quad\int_{-\infty}^{b}f(x)\,dx=\lim_{a\to-\infty}\int_{a}^{b}f(x)\,dx,\]

Figure 1.48: Intermediate Value Theorem for integrals

provided that the limits exist. For example,

\[\int_{1}^{\infty}x^{-2}dx=\lim_{b\to\infty}\int_{1}^{b}x^{-2}dx=\lim_{b\to\infty} \left.\frac{-1}{x}\right|_{1}^{b}=\lim_{b\to\infty}\left(1-1/b\right)=1.\]

Similarly, integrals with a singularity at an end point of integration can be computed. For example, if \(f\) is defined in \((a,b]\) and is bounded and integrable on each subinterval \([c,b]\) for \(a

\[\int_{a}^{b}f(x)\,dx=\lim_{c\to a^{+}}\int_{c}^{b}f(x)\,dx,\]

if the limit exists. For example, suppose we have \(\int_{0}^{1}\frac{1}{x}\,dx\). Then for each \(c\in(0,1)\), we have \(\int_{c}^{1}\frac{1}{x}\ dx=\ln 1-\ln c=-\ln c\). Since \(\lim_{c\to 0^{+}}(-\ln c)=\infty\), the given integral \(\int_{0}^{1}\frac{1}{x}\ dx\) diverges. On the other hand, \(f(x)=\frac{1}{\sqrt{x}}\) is still undefined at \(x=0\) but

\[\int_{0}^{1}\frac{1}{\sqrt{x}}\ dx=\lim_{c\to 0^{+}}2\sqrt{x}\Bigg{|}_{c}^{1}=2.\]

### 18 The Fundamental Theorem of Calculus

The derivative and integral are defined independently. The definition of derivative is motivated by the problem of finding slope of the tangent line and is expressed as

\[f^{\prime}(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}.\]

It is a functional limit of difference quotients. The definition of integral is motivated by finding areas under non-constant functions and is given in terms of supremums and infimums of finite sums:

\[\int_{a}^{b}f(x)\,dx=U(f)=L(f).\]

The Fundamental Theorem of Calculus reveals a remarkable relationship between these two concepts, which says that integration and differentiation are inverse processes. The Fundamental Theorem of Calculus is really two theorems. The first one states under some conditions on the function “the derivative of an integral of a function is the original function.” The second result establishes the reverse “the integral of the derivative of a function is again the same function.” We start with a theorem about differentiation of the integral.

**Theorem 51** (Fundamental Theorem of Calculus I).: _Suppose that \(f\) is an integrable function on \([a,b]\), and define_

\[F(x)=\int_{a}^{x}f(t)\,dt,\quad a\leq x\leq b.\]

_Then \(F\) is continuous on \([a,b]\). Furthermore, \(F\) is differentiable at each point \(x_{0}\in(a,b)\), where \(f\) is continuous, and \(F^{\prime}(x_{0})=f(x_{0})\)._

Proof

: By definition \(f\) is bounded. Let \(M>0\) be such that \(|f(x)|\leq M\) for all \(x\in[a,b]\). Take \(x,y\in[a,b]\), and observe that

\[|F(y)-F(x)|=\left|\int_{x}^{y}f(t)\,dt\right|\leq M|x-y|.\]

This shows that \(F\) is (uniformly) continuous on \([a,b]\). Now let us assume that \(f\) is continuous at \(x_{0}\in[a,b]\). In order to show \(F^{\prime}(x_{0})=f(x_{0})\), write \(F^{\prime}(x_{0})\) as

\[\lim_{x\to x_{0}}\frac{F(x)-F(x_{0})}{x-x_{0}} =\lim_{x\to x_{0}}\frac{1}{x-x_{0}}\left(\int_{a}^{x}f(t)\,dt- \int_{a}^{x_{0}}f(t)\,dt\right)\] \[=\lim_{x\to x_{0}}\frac{1}{x-x_{0}}\left(\int_{x_{0}}^{x}f(t)\,dt\right)\]

and thus

\[\frac{F(x)-F(x_{0})}{x-x_{0}}-f(x_{0})=\frac{1}{x-x_{0}}\int_{x_{0}}^{x}[f(t) -f(x_{0})]dt,\quad x\in(a,b),x\neq x_{0}.\]